Wow... this essentially proved that if you take the product of four consecutive -numbers- integers and add one to it, than it's gone be a square number.
My professor had us prove a more general result: take the product of four numbers in arithmetic sequence, then add the fourth power of their common difference. Show that the result is a perfect square.
@@leif1075 well it took me about 5min to solve it so I think is not impossible to solve. All of this types of equations where you have 4 consecutive numbers multipled are done like this
Lol,This problem is actually super easy,(Every single Olympiad contestant would have solved this question,at some point of their life) .Insanely hard problems need not have simple solutions . That's a downside of the math Olympiad .They make you expect difficult problems have simple solutions.(Although,most imo contestants don t fall for this fallacy).Real insanely hard problems have not been solved by anyone,yet.
That's pretty funnt, but sometimes such things are the most difficult to see, for example: we have f(x)=x^4+8x^3+18x^2+8x+17, and a question, for which x, the function f(x) is a prime. You can check infinitely many cases and never know the answer, but what makes this question easy (but on the other hand is not so obvious), is that 18=17+1. Because then we have (x^2+1)(x^2+8x+17), which has to be a prime. One of them has to be = 1, the other one has to be some prime then... We are left with only 2 cases, because we know, that 18=17+1. : )
Not only 2 = 1+1, but also 0 = 1-1. From the second row: (x^2+3x+1-1)(x^2+3x+1+1)+1 and per the third binomial equation = (x^2+3x+1)^2 -1^2 +1 = (x^2+3x+1)^2
@@yogeshpathak73 I think Tibor Grün is from Germany. In German school curriculum the formula (a+b)*(a-b) = a^2 - b^2 is known as 3. binomial formula. b=1 is a special case.
Nice. I did it this way: Assume that the expression is a square number so: x(x+1)(x+2)(x+3)+1 = n^2 x(x+1)(x+2)(x+3) = n^2 - 1 x(x+1)(x+2)(x+3) = (n+1)(n-1) What I did then is realise that the factors of the product on the right differ by 2. Playing around you can find that: x(x+3) = x^2+3x = n-1 (x+1)(x+2)=x^2+3x+2 = n+1 So n = x^2 + 3x + 1 Not as neat as your method though! Thanks for the video
@@joshuamason2227 Well you have the product of 3 binomials and a monomial for which we can multiply in any order. If you try a few cases, or think about it you spot that x(x+3) and (x+1)(x+2) have a difference of 2.
3:20 I solved it differently. Let y= x^2+3x. Then substitute y into the expression making y(y+2)+1, distribute so y^2+2y+1 and that is a perfect square of (y+1)^2. Here, the square root and exponent cancel each other leaving y+1, sub back in x and then easily find the answer :)
I expressed it as sqrt((501.5-1.5)(501.5-0.5)(501.5+0.5)(501.5+1.5)+1) You get two a^2-b^2 expressions that you can multiply out, add the 1, and then factor into a squared quadratic expression Very neat and, as someone mentioned elsewhere, it generalizes to “1 plus the product of any four consecutive integers is a perfect square”
I chose to make x = 502, which ends up yielding a nice difference of squares and a two term quadratic, which is much easier to distribute. The quartic you get has a palindromic pattern reminiscent of pure binomial coefficients, making it tempting to say the golden ratio is a root. It is, in fact, a root, so synthetically divide the quartic by the golden ratio identifying polynomial, x² - x - 1. You end up with the golden ratio identifying polynomial again, meaning that the original quartic in that square root is (x² - x - 1)², so cancel the power and the root. Plug 502 back in for x, some quick multiplying and subtracting by hand and you've got 251501.
I'm only in 8th grade Algebra 1 but I was using variables to find how some of your factorizations works. You went from (x^2+3x)(x^2+3x+2)+1 to (x^2+3x)(x^2+3x+1)+(x^2+3x+1). What I did was set (x^2+3x) to a variable (a). (a)(a+2)+1 a^2+2a+1 (a+1)(a+1) Now substitute back in. (x^2+3x+1)(x^2+3x+1) When in doubt use variables..
Nice factoring method but it might have taken me a while to spot. Multiplying out and factoring isn't so bad (x - 1)x(x + 1)(x + 2) + 1 = (x^2 - 1)(x^2 + 2x) + 1 = x^4 + 2x^3 - x^2 - 2x + 1 = (x^2 + bx +- 1)^2 = x^4 + 2bx^3 + (b^2 +- 2)x^2 +- 2bx + 1 We see this works if b = 1 and c = -1 so the answer is 501^2 + 501 - 1 = 500^2 + 2*500 + 1 + 500 = 251501
Another nice solution is to assume x=501.5 And rewrite the equation which would give x⁴-(5/2) x²+(9/16) +1 which is basically (x²-5/4) ² The square and square root will cancel and give x²-5/4 Taking lcm would give us ((2x)² - 5) /4 (2x)²=1003² which can be computed very easily as 1003=1000+3 And then we just have to subtract 5 and divide by 4
You can also this as x^2+3x=t and expression would become t(t+2)+1 =(t+1)^2 this que came in practice test for jee last week And guess what i solved that 😎😎😎👍👍
I did assume there was a nice solution, but expanding under the root to get x^4 + 6x^3 + 11x^2 + 6x + 1 was pretty easy, and then matching coefficients in (x^2 + ax + 1)^2 was straightforward too. But yeah, the main thing is to replace 500 by x. I don't think I could intuitively see which two of the brackets would make it easier, and I'm not sure that's a better method than expanding the whole thing to only 4 terms (plus the one on the outside).
Just wanted to say after some work, some variable assigning and a lucky coincidence later, I found the answer! My steps: Let a=500 Expand a(a+1)(a+2)(a+3) to a^4+6a^3+11a^2+6a+1 Complete the square (or the fourth in this case): a^4+4a^3+6a^2+4a+1+2a^3+5a^2+2a =(a+1)^4+2a(a^2+2a+1+.5a) =(a+1)^4+2a((a+1)^2+.5a) =(a+1)^4+2a(a+1)^2+a^2 Observe this follows the perfect square structure. Therefore: (a+1)^4+2a(a+1)^2+a^2 =[(a+1)^2+a]^2 Square rooting gives: (a+1)^2+a a^2+3a+1 By substitution: a^2+3a+1 =250000+1500+1 =251501
I love the explanation, though I did it a bit differently. When I got to the second line, I substituted (x²+3x) as y and found that that worked much simpler than distributing 2 as 1+1.
I got the point. Convert the number into variables. For example 500=x or 2=1+1 = a (convert what you want) Thank you for the useful tips... I realized that the algebra is so amazing at the complicated situation.
The factorization was more if you defined a variable "a" that was equal to x^2+3x Because since you multiply and you have left (X^2+3X)(X^2+3X+2)+1 With the variable "a" you had left (a)(a+2)+1 And that is equal to (a^2+2a+1) And that is factorizable as (a+1)^2 Greetings from Mexico
Actually why not: √(500)(501)(502)(503)+1 = x 500.501.502.503 +1 = x² 500.501.502.503 = x²-1 (500.503) (501.502) =(x-1)(x+1) 251500 × 251502 = (x-1) (x+1) x = ±251501 but we have to reject -251501 because it is negative in a square root
If somebody can't understand 3:59, you can try another method; you can also substitute the line 3 of 'Obs' into another variable, for example: y If: y = x²+3x Then the equation becomes: = (y+2)(y)+1 = y²+2y+1 Factorize that into this: = (y+1)² So, sqrt[(y+1)²] = y+1 Since y = x²+3x The equation becomes: = x²+3x+1 Since x = 500 The result is: = (500)²+3(500)+1 = 251501
Solved it! For a clean solution to exist I assumed that x·(x + 1)·(x + 2)·(x + 3) + 1 is a perfect square for any integer x. Playing around with x = 1 & x = 2 it is quickly apparent that x·(x + 3) + 1 is a contender for the solution. It is easy to prove that this is the solution by expanding out [x·(x + 3) + 1]² and showing that it is equivalent to x·(x + 1)·(x + 2)·(x + 3) + 1 I doubt I would have spotted the algebraic manipulation that BPRP used without knowing the solution first. I also learn something new i.e. the product of four consecutive integers plus 1 is always a perfect square. Thank you for the video - I enjoyed this one.
I'm doing this on the toilet, so I only hope I'm starting correctly, with (500)(502)=(501²-1) and (501)(503)=(502²-1) But then again, we could cheat and go with (501)(502)=(501½²-¼) and (500)(503)=(501½²-9/4)?
You could also do like: Consider 501 as “x” and 502 as “y” You can rewrite the sentence like: (x-1).(x+1).(y-1).(y+1) +1 That’s equal to: (x^2 - 1^2).(y^2 - 1^2) +1 Or (501^2 - 1).(502^2 - 1) +1 And there’s your answer xD!!
I love your videos about not using calculators (Like the Wolfram-Alpha video)! They're the best! Keep up the good work! It's nice going back to algebra sometimes...
If you guys want to know the “secret”, it is if the sum of the digits in any number add up to a multiple of 3, it is divisible by 3. E.g 567: 5+6+7=18. 18 is a multiple of 3 thus 567 is divisible by 3.
Hmmm ive got an easier way when you use 1 + 1 instead of 2 Here is how I do according to you: (x^2+3x+2)(x^2+3x+1) =(x^2+3x)^2 +2 then √((x^2+3x)^2 +2) = x^2+3x+1
I did it by recognizing that (x^2+3x)(x^2+3x+2) is easily simplified with a substitution of y=(x^2+3x+1). It simplifies to (y-1)(y+1) = y^2 - 1. Since we have a "+1" hanging out after the 4-term product, that gets rid of the "-1" in our simplified expression, yielding just y^2 under the radical sign. square root of y^2 = y. That means the solution is our substitution: y=(x^2+3x+1). Plugging in 500 for x gives us 251501.
I know this is not related to this video but I wanted to post this on a new video so you might see it :) your trick for integrals of thinking "wouldn't it be nice if..." has helped me so so much, so thank you :) love your videos!
Your way to solve this is pretty AWESOME!! First I multiplied all together, i got x^4 + 6x^3 + 11x^2 + 6x + 1 then i calculate this polynomial for x=1 x=2 x=3 ....all the time i got a square!! I was really surprise!! I didn't expect x(x+1)(x+2)(x+3)+1 to be a perfect square for all x at all!!! This is incredible!! Thanks for sharing your knowledge you are very inspiring to me
I find the solution in a different way: k(k+1)(k+2)(k+3)=x^2-1 =(x-1)(x-2), and the difference between these number is 2; so multiplying k(k+3)=k^2+3k (k+1)(k+2)=k^2+3k+2 I have the numbers with desidered difference. So x=k(k+3)+1, having the result with substitution 500->k.
if only most teachers were like this guy, it actually makes me wanna learn math again and I'm 29 years old! not gonna lie that did look fun for some reason.
If you guys want a tip to make the solution simpler Set 501 as x instead of 500 That way (x-1)(x)(x+1)(x+2)+1 is under the square root Which many can see the only tedious multiplying we’d have to do is (x^2 - 1)(x^2 + 2) You’ll then find the equation to be x^2 + x - 1 Which gives the same result Just helps save space and makes the problem less tedious
Dude, I always had a good grasp on algebra as a kid and in highschool I always aced most algebra, but somehow my teachers (and I) missed this property of algebraic equations. So freaking cool. It has been nigh on 15 years since high school, but I am still learning new and cool algebra. Thanks so much blackpenredpen!
what about using x. First allow the equation equal to x, then square both sides and then both sides minus 1. We get (x+1)(x-1)=(500)(501)(502)(503). Then I found out that (500)(503)is two less than (502)(501) and the answer would be 500*503+1=251501. The last step can be done be simple calculation, no calculator needed.
I multiplied everything out and got stuck... This is a brilliant solution. One day I will achieve this type of mathematical intuition. Lead the way, blackpenredpen!!!
I think you forgot about the absolute value; Square root of a 2nd power produces absolute value result because both positive and negative values are true.
You could have made it simpler using the (a-b)(a+b)=a^2-b^2 formula. (x^2+3x)(x^2+3x+2)+1 = (x^2+3x+1-1)(x^2+3x+1+1)+1 = (x^2+3x+1)^2-1^2+1 = (x^2+3x+1)^2
Wow your method and my method are similar .....I had take the whole expression as y and then square it and then assume x to be 500 and multiplied and had taken x^2+3x to be z and at end I got y=z+1 that is y = x^2 + 3x + 1 and it's done
At 3:45, you could also treat the first factor as (x^2+3x+1-1) so together with the second factor you have (x^2+3x+1-1)(x^2+3x+1+1) = difference of squares ((x^2+3x+1)^2 - 1. Plus the extra 1 on the outside you get the perfect square.
the easier way would be to solve it symmetrically: lets say x equals 501,5 in this case.. then the product would be (x-1,5)(x+1,5)(x-0,5)(x+0,5)+1= (x^2-2,25)(x^2-0,25)+1= x^4-2,5*x^2+0,5625+1 which is obviously (x^2-1,25)^2
When a number repeats itself in an addition you can factor it out, basically do the inverse of distributive property. So we have x²+3x+1 repeating in both therms. You can factor it out and you will be left with x²+3x+1 ( x²+3x +1), equivalent to x²+3x+1( x²+3x) + x²+3x+1 (1)
i did some mental math, but hit a wall at trying to find the square root of 63,252,753,001
that’s some impressive mental math assuming you’re telling the truth . Is there a trick or something
@@iamgroot3615 theres no trick hes probably lying
r/iamverysmart
AngryAxew there's no reason not to be able to mental math those numbers
Like 500(500+1) which is easier which is 250000+500 and it similar to the end
I mean it's OBVIOUSLY 251501.
I just started learning English, but the explanations are clear and interesting even at my levels of English. Thanks a lot 😁👍
Юлия Охременко I am
Glad to hear!
x2
U r indian Chinese korean or ....???
@@harelavv8806 the name may seem obviously Russian to some but not all
@@blackpenredpen hii
Wow... this essentially proved that if you take the product of four consecutive -numbers- integers and add one to it, than it's gone be a square number.
Awesome! Good spot!
Gábor Tóth holy shit you’re right! That’s crazy man
Yup!!
The most pathological case I can think of is -1 thru 2, and yes indeed I get 1, which is a perfect square.
My professor had us prove a more general result: take the product of four numbers in arithmetic sequence, then add the fourth power of their common difference. Show that the result is a perfect square.
Instead distributing at 4:18
u = x^2 + 3x + 1
(u - 1)(u + 1) + 1 = u^2
So the root is x^2 + 3x + 1 = 251501
Bryan Lu omg that cat!!!!
Or even u=x^2+3x, then u^2+2u+1
digno de nyan cat, jajajajaj
My life is a lie. I thought u subbing was only for integrals
@@mattat3847 nah man, sub whenever it makes the problem simpler
Olympic math taught me that insanely hard problems often had elegant solutions, this is no exception.
: ))))
@@leif1075???
@@leif1075 people like these are called problem solvers...
@@leif1075 well it took me about 5min to solve it so I think is not impossible to solve. All of this types of equations where you have 4 consecutive numbers multipled are done like this
Lol,This problem is actually super easy,(Every single Olympiad contestant would have solved this question,at some point of their life) .Insanely hard problems need not have simple solutions . That's a downside of the math Olympiad .They make you expect difficult problems have simple solutions.(Although,most imo contestants don t fall for this fallacy).Real insanely hard problems have not been solved by anyone,yet.
1990: we'll have flying cars by 2019
2019: 2=1+1, wow I'm a genius
LOL
2+2 is 4, minus one that's three quick maths
Flying cars... you can't even have a sharpie that could change color. Say, red and black.
@@ghotifish1838 topical meme reference
Well it can also be 2=500-498
My last words whispered in a final breath : "Don't forget the +1"
😂
LMFAOOO
He never resumed the video
He added the one wdym 🧐
i put it in my calculator and got 251501, that was easy
Boo!
@The Balton American calculators are beasts
@@Netherexio Agreed but Americans aren’t
@@paradox9265 What do you mean?
It was easy, but not so beautiful like this)
Did you know that 2 = 1 + 1?? I bet not!
jk : )
blackpenredpen No i don’t, i need a calculator to this
That's pretty funnt, but sometimes such things are the most difficult to see, for example: we have f(x)=x^4+8x^3+18x^2+8x+17, and a question, for which x, the function f(x) is a prime. You can check infinitely many cases and never know the answer, but what makes this question easy (but on the other hand is not so obvious), is that 18=17+1. Because then we have (x^2+1)(x^2+8x+17), which has to be a prime. One of them has to be = 1, the other one has to be some prime then... We are left with only 2 cases, because we know, that 18=17+1. : )
No but I knew 2= 0.9+1.1.
I Known 2+2 = 5
What i know is 5/2 = 2 with int data type
Dafuq did i jusf watch.i lost it when the 2=1+1
You are not nerdy enough
no 2=1+1/2+1/4+1/8+1/16...
you have got many misconceptions blackpenredpen!!!
When he writes 1, he's obviously just abbreviating 1/2+1/4+1/8+1/16+...
@@iabervon and when he writes 1/2 he's abbreviating for 1/4+1/8+1/16+...
@@InDstructR And when he writes 1/4 he's abbreviating 1/8 + 1/16 + 1/32+...
@ki kus won't stop me,
And when he writes 1/8 he's abbreviating 1/16+1/32+1/64+1/128+...
Whoa that converged quickly
I remember solving this exact question in my JEE ( Mains ) exam.
@Sanat R mains usually has easy questions
@Sanat R - Study Vlogs Sure, yeah, "easy question" 😬
@Sanat R - Study Vlogs Woah, really? What kinda questions do they ask? Could you send me a link?
1 + 1 = 3
And sinx/n=six=6.
You stupid, 1 + 1 ≠ 3
3 = 2 + 1
π = 2 + 1
π - 1 = 2
@@rio_agustian_ so π = 3 lol nice discovery
@@CookieJar2025 e = 3 = π
@@rio_agustian_ noob
@@Kevin-14 lool nooob
"If you're using a calculator, why are you watching this video?"
Sanity check.
hahahahaa
I am a calculator, not a person
*laughs in Shakuntala Devi
Take x ^2 + 3x = a
Then in step 2
a(a+2) + 1
a^2 + 2a + 1
= (a+1)^2
Yes much easier, and you see it imediatly too.
was about to say this, i think it's a lot more intuitive
Isn't it beautiful how one problem can be solved in diffrent ways, even if the idea and the method are nearly the same. That's why we love maths.
@@milanmitreski7657 Yes
You extra smart boy the time required here will be same
Not only 2 = 1+1, but also 0 = 1-1.
From the second row:
(x^2+3x+1-1)(x^2+3x+1+1)+1 and per the third binomial equation
= (x^2+3x+1)^2 -1^2 +1
= (x^2+3x+1)^2
I didn't understand. Where did you use binomial expansion
0 = 1-1
1 = 1*1
2= 1+1
I didn't see any binomial here... But what i see is that you used the form (a+1)(a-1) + 1 = a^2 - 1 +1= a^2
@@yogeshpathak73 I think Tibor Grün is from Germany. In German school curriculum the formula (a+b)*(a-b) = a^2 - b^2 is known as 3. binomial formula. b=1 is a special case.
Oh ok... Didn't know that. Thanks.
This is beautiful. I've been looking at it for five hours now
Nice. I did it this way:
Assume that the expression is a square number so:
x(x+1)(x+2)(x+3)+1 = n^2
x(x+1)(x+2)(x+3) = n^2 - 1
x(x+1)(x+2)(x+3) = (n+1)(n-1)
What I did then is realise that the factors of the product on the right differ by 2. Playing around you can find that:
x(x+3) = x^2+3x = n-1
(x+1)(x+2)=x^2+3x+2 = n+1
So n = x^2 + 3x + 1
Not as neat as your method though!
Thanks for the video
OMG ive almost done it completely. i just stopped at (n+1)(n-1) lol WD! i mean 'not that almost' lmao
How do I play around with it
@@joshuamason2227 Well you have the product of 3 binomials and a monomial for which we can multiply in any order. If you try a few cases, or think about it you spot that x(x+3) and (x+1)(x+2) have a difference of 2.
@@dr3w199 okie
Damn... It's a great method. Neat work. 💯
0:09 That was PowerFul
Continueing from: sqrt( (x^2 + 3x) (x^2 + 3x + 2) + 1 )
let y = x^2 + 3x
sqrt( y * (y + 1) + 1 )
= sqrt( y^2 + y + 1 )
= sqrt( (y+1)^2 )
= y + 1
= x^2 + 3x + 1
= (x + 1) (x + 2) - 1
= 501 * 502 - 1
= 251501
much easier to multiply :p
that is what i wanted ti type nice 👍
0:10 is this a pewdiepie reference? 😂😂
He liked it!!!
@@albel2094 yupp
Yupp
3:20 I solved it differently.
Let y= x^2+3x. Then substitute y into the expression making y(y+2)+1, distribute so y^2+2y+1 and that is a perfect square of (y+1)^2.
Here, the square root and exponent cancel each other leaving y+1, sub back in x and then easily find the answer :)
this is also what I did and I think that this is a bit better because you don't have to split 2 into 1 + 1 and do the rest
That's what I did as well
Jeez thats smart
*proceeds to use the calculator to prove that 251501 is the right answer*
I expressed it as sqrt((501.5-1.5)(501.5-0.5)(501.5+0.5)(501.5+1.5)+1)
You get two a^2-b^2 expressions that you can multiply out, add the 1, and then factor into a squared quadratic expression
Very neat and, as someone mentioned elsewhere, it generalizes to “1 plus the product of any four consecutive integers is a perfect square”
You can also put a +1-1 inside the x^2+3x bracket and it'll be in the form of (a+b)(a-b).
that's what i thought he was gonna do as well but what he did was cool as well.
why?
Yeah, (x-1)(x+1)+1=x^2-1^2+1 seems easier to find than multiplying out exactly the right portion of the big expression.
I chose to make x = 502, which ends up yielding a nice difference of squares and a two term quadratic, which is much easier to distribute. The quartic you get has a palindromic pattern reminiscent of pure binomial coefficients, making it tempting to say the golden ratio is a root. It is, in fact, a root, so synthetically divide the quartic by the golden ratio identifying polynomial, x² - x - 1. You end up with the golden ratio identifying polynomial again, meaning that the original quartic in that square root is (x² - x - 1)², so cancel the power and the root. Plug 502 back in for x, some quick multiplying and subtracting by hand and you've got 251501.
I'm only in 8th grade Algebra 1 but I was using variables to find how some of your factorizations works.
You went from (x^2+3x)(x^2+3x+2)+1 to (x^2+3x)(x^2+3x+1)+(x^2+3x+1).
What I did was set (x^2+3x) to a variable (a).
(a)(a+2)+1
a^2+2a+1
(a+1)(a+1)
Now substitute back in.
(x^2+3x+1)(x^2+3x+1)
When in doubt use variables..
Yes, that's using even more algebra than BPRP did.
Well yes because using the variables is actually the logic behind the solution, it's just that it was invisible throughout the process :D
Where r u from?
Agreed! Variables always help to proceed the solution.
I put x=500 but multiplied everything. In the end i got to sqrt((x+y)^2) with x=500 and y=501^2
5:48 “Back in my day kids would use *ALGEBRA* but now their brains are rotting from these darn *CALCULATORS* ”
3:20 just put y = x^2 + 3x, then you have y(y+2) + 1 = y^2 + 2y + 1 = (y+1)^2. So the answer is y + 1, or x^2 + 3x + 1.
I also did it in this way...but that way was also fine...its all about which method comes in your head first
Dang that’s genius
Seemingly elementary problems can have wonderfully elegant solutions! All we need is to substitute a number with x, and the magic begins.
i remember my math teacher asking me to prove that n(n+1)(n+2)(n+3) + 1 is always a perfect square given that n is an integer
The world needs more teachers like you. I'm more impressed by your teaching skills than any math. Much respect.
0:01 that's my life philosophy now
i watched this video this video right before my math competition and the same type of question came up on the task sheet. Thank you very much!
for those wondering the question was
202120212019(202120212021)(202120212023) all over 100010001 x (202120212021 squared +4)
now do it with CALCULUS
Why do you think this is funny?
@@davidappell3105 because suffering is funny
@@davidappell3105 Its FUNI
Sorry...Time over! give me your exam!
Nice factoring method but it might have taken me a while to spot. Multiplying out and factoring isn't so bad
(x - 1)x(x + 1)(x + 2) + 1 = (x^2 - 1)(x^2 + 2x) + 1 = x^4 + 2x^3 - x^2 - 2x + 1 = (x^2 + bx +- 1)^2
= x^4 + 2bx^3 + (b^2 +- 2)x^2 +- 2bx + 1
We see this works if b = 1 and c = -1 so the answer is 501^2 + 501 - 1 = 500^2 + 2*500 + 1 + 500 = 251501
Another nice solution is to assume x=501.5
And rewrite the equation which would give
x⁴-(5/2) x²+(9/16) +1 which is basically (x²-5/4) ²
The square and square root will cancel and give x²-5/4
Taking lcm would give us
((2x)² - 5) /4
(2x)²=1003² which can be computed very easily as 1003=1000+3
And then we just have to subtract 5 and divide by 4
You can also this as x^2+3x=t and expression would become t(t+2)+1 =(t+1)^2 this que came in practice test for jee last week And guess what i solved that 😎😎😎👍👍
Why are your videos so entertaining? I'm so glad I came across this channel.
I am so impressed with myself, I actually used the same method you did before watching the video =D
Jan Wrobel nice!!!!
When he drops the "Check this out", you know crazy stuff will happen on the board
Every body knows 1+1=2 but i know 1+1 =/= 3
♫♪Ludwig van Beethoven♪♫ Hahahhaha
I’ve got you all beat with 1+1 > 0
@@jgsh8062 nah mine's better 1+1≠1+1
Doing a PhD in Literary Studies, but stuff like this is why I absolutely love maths ♥
What a incredible content. Im a student of math (i'll be a teacher in the future) from Brazil. Thank you so much for sharing knowledge!
Your explanation is awesome . I like your teaching very much. Thanks
I did assume there was a nice solution, but expanding under the root to get x^4 + 6x^3 + 11x^2 + 6x + 1 was pretty easy, and then matching coefficients in (x^2 + ax + 1)^2 was straightforward too.
But yeah, the main thing is to replace 500 by x. I don't think I could intuitively see which two of the brackets would make it easier, and I'm not sure that's a better method than expanding the whole thing to only 4 terms (plus the one on the outside).
3:43
If we assume x^2+3x to be t we get t + 1 whole squared which is a lot easier
I know this kind of the prob, i use (n+1)(n+2)-1
Just wanted to say after some work, some variable assigning and a lucky coincidence later, I found the answer!
My steps:
Let a=500
Expand a(a+1)(a+2)(a+3) to a^4+6a^3+11a^2+6a+1
Complete the square (or the fourth in this case):
a^4+4a^3+6a^2+4a+1+2a^3+5a^2+2a
=(a+1)^4+2a(a^2+2a+1+.5a)
=(a+1)^4+2a((a+1)^2+.5a)
=(a+1)^4+2a(a+1)^2+a^2
Observe this follows the perfect square structure.
Therefore:
(a+1)^4+2a(a+1)^2+a^2
=[(a+1)^2+a]^2
Square rooting gives:
(a+1)^2+a
a^2+3a+1
By substitution:
a^2+3a+1
=250000+1500+1
=251501
I love the explanation, though I did it a bit differently. When I got to the second line, I substituted (x²+3x) as y and found that that worked much simpler than distributing 2 as 1+1.
Blew my mind! Earned yourself a new subscriber! Keep up the good work!👍
Now this video makes me like algebra
th-cam.com/video/XX2DI9E1zV8/w-d-xo.html
I got the point. Convert the number into variables. For example 500=x or 2=1+1 = a (convert what you want)
Thank you for the useful tips... I realized that the algebra is so amazing at the complicated situation.
Really good solution! GOOD Teacher👍
a very good perspective and a very good solution. thank you!!!
this guy is a genius!
The factorization was more if you defined a variable "a" that was equal to x^2+3x
Because since you multiply and you have left
(X^2+3X)(X^2+3X+2)+1
With the variable "a" you had left
(a)(a+2)+1
And that is equal to (a^2+2a+1)
And that is factorizable as (a+1)^2
Greetings from Mexico
What Everybody knows : 1+1=2
What BPRP knows : 2=1+1
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.
.
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What I know : 1+1=2 and 2=1+1
😇😇😇😇😇😇😇😇😇😇😇
Actually why not:
√(500)(501)(502)(503)+1 = x
500.501.502.503 +1 = x²
500.501.502.503 = x²-1
(500.503) (501.502) =(x-1)(x+1)
251500 × 251502 = (x-1) (x+1)
x = ±251501 but we have to reject -251501 because it is negative in a square root
Everybody knows e^{iτ}=1
.
.
.
.
But I know 1=e^{iτ}
Nice!!!
@@日本人じゃありません That's e^{iπ}. τ=2π
My mans using tau! Up top!
@@日本人じゃありません he's using tau not pi
If somebody can't understand 3:59, you can try another method; you can also substitute the line 3 of 'Obs' into another variable, for example: y
If: y = x²+3x
Then the equation becomes:
= (y+2)(y)+1
= y²+2y+1
Factorize that into this:
= (y+1)²
So, sqrt[(y+1)²] = y+1
Since y = x²+3x
The equation becomes:
= x²+3x+1
Since x = 500
The result is:
= (500)²+3(500)+1
= 251501
I have never been so hyped at 2 = 1+1 before.
Solved it!
For a clean solution to exist I assumed that
x·(x + 1)·(x + 2)·(x + 3) + 1 is a perfect square for any integer x.
Playing around with x = 1 & x = 2 it is quickly apparent that
x·(x + 3) + 1 is a contender for the solution.
It is easy to prove that this is the solution by expanding out
[x·(x + 3) + 1]² and showing that it is equivalent to x·(x + 1)·(x + 2)·(x + 3) + 1
I doubt I would have spotted the algebraic manipulation that BPRP used without knowing the solution first. I also learn something new i.e. the product of four consecutive integers plus 1 is always a perfect square.
Thank you for the video - I enjoyed this one.
I'm doing this on the toilet, so I only hope I'm starting correctly, with (500)(502)=(501²-1) and (501)(503)=(502²-1)
But then again, we could cheat and go with (501)(502)=(501½²-¼) and (500)(503)=(501½²-9/4)?
you can also use substitution x^2+3x = t and you get t(t+2)+1=t^2+2t+1=(t+1)^2 and replace t: (x^2+3x+1)^2
You could also do like:
Consider 501 as “x” and 502 as “y”
You can rewrite the sentence like:
(x-1).(x+1).(y-1).(y+1) +1
That’s equal to:
(x^2 - 1^2).(y^2 - 1^2) +1
Or
(501^2 - 1).(502^2 - 1) +1
And there’s your answer xD!!
A generalisation of the algebraic expression - (X)(X+1)(X+2)(X+2)+1= (Y)^2
I love your videos about not using calculators (Like the Wolfram-Alpha video)! They're the best! Keep up the good work! It's nice going back to algebra sometimes...
Why It?
Yea me too. I try to mix things up a bit.
If you guys want to know the “secret”, it is if the sum of the digits in any number add up to a multiple of 3, it is divisible by 3.
E.g 567: 5+6+7=18. 18 is a multiple of 3 thus 567 is divisible by 3.
@Cerebro Spinal I was referring to the part at the end of the video
Hmmm ive got an easier way when you use 1 + 1 instead of 2
Here is how I do according to you:
(x^2+3x+2)(x^2+3x+1)
=(x^2+3x)^2 +2 then
√((x^2+3x)^2 +2) = x^2+3x+1
You can't solve squareroots of sums like that, eventhought your result is correct.
I did it by recognizing that (x^2+3x)(x^2+3x+2) is easily simplified with a substitution of y=(x^2+3x+1). It simplifies to (y-1)(y+1) = y^2 - 1. Since we have a "+1" hanging out after the 4-term product, that gets rid of the "-1" in our simplified expression, yielding just y^2 under the radical sign. square root of y^2 = y. That means the solution is our substitution: y=(x^2+3x+1). Plugging in 500 for x gives us 251501.
I know this is not related to this video but I wanted to post this on a new video so you might see it :) your trick for integrals of thinking "wouldn't it be nice if..." has helped me so so much, so thank you :) love your videos!
Awww thank you!!!!!
Your way to solve this is pretty AWESOME!! First I multiplied all together, i got x^4 + 6x^3 + 11x^2 + 6x + 1 then i calculate this polynomial for x=1 x=2 x=3 ....all the time i got a square!! I was really surprise!! I didn't expect x(x+1)(x+2)(x+3)+1 to be a perfect square for all x at all!!! This is incredible!! Thanks for sharing your knowledge you are very inspiring to me
There are things to learn from each of your videos 😁❤️
Well yes, that's the point.
I find the solution in a different way:
k(k+1)(k+2)(k+3)=x^2-1
=(x-1)(x-2),
and the difference between these number is 2;
so multiplying
k(k+3)=k^2+3k
(k+1)(k+2)=k^2+3k+2
I have the numbers with desidered difference.
So x=k(k+3)+1, having the result with substitution 500->k.
please give an example differentiation of complex functions
An alternative way to solve it is by letting x = 501.5 and change the expression into
sqrt((x - 3/2)(x - 1/2)(x + 1/2)(x + 3/2) + 1)
The inspiration for this is difference of squares, simplifying gives.
= sqrt( (x^2 - 9/4)(x^2 - 1/4) + 1)
= sqrt( x^4 - 10/4(x^2) + 9/16 + 1)
= sqrt(x^4 - 5/2(x^2) + 25/16)
= sqrt((x^2 - 5/4)^2) , *factors nicely, perfect square*
= x^2 - 5/4
= (501.5)^2 - 1.25
= 251501
Except the 501.5^2 part is not so pleasant by hand. Not the best alternative.
BPRP know 2 = 1+1
I know 2 = 2
what happened to the comment button its gray
I know 2 = two
if only most teachers were like this guy, it actually makes me wanna learn math again and I'm 29 years old! not gonna lie that did look fun for some reason.
"And now, here's the deal"… You know that when he pronounces that phrase things are 'bout to get complicated.
If you guys want a tip to make the solution simpler
Set 501 as x instead of 500
That way (x-1)(x)(x+1)(x+2)+1 is under the square root
Which many can see the only tedious multiplying we’d have to do is (x^2 - 1)(x^2 + 2)
You’ll then find the equation to be
x^2 + x - 1
Which gives the same result
Just helps save space and makes the problem less tedious
Everybody know e^2.pi.i = 1
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But I know 1 = e^2.pi.i
Wrong it's
- (e ^ pi × i)
@@mundane3809 Nice try but thats -1
ignoring your name
@@nikolas9105 no
e ^ ( pi × i ) = -1
So if you make -1 negative, it become positive.
@@mundane3809 you are correct but the original comment was also correct. e^2πi = 1.
@@RunstarHomer oof yea it's actually correct. sorry for the mistake!
More impressed with how someone came up with the question
Dude, I always had a good grasp on algebra as a kid and in highschool I always aced most algebra, but somehow my teachers (and I) missed this property of algebraic equations. So freaking cool. It has been nigh on 15 years since high school, but I am still learning new and cool algebra. Thanks so much blackpenredpen!
what about using x.
First allow the equation equal to x, then square both sides and then both sides minus 1. We get (x+1)(x-1)=(500)(501)(502)(503). Then I found out that (500)(503)is two less than (502)(501) and the answer would be 500*503+1=251501. The last step can be done be simple calculation, no calculator needed.
Shout-out to my colorblind fam who can never tell when he switches pens
I multiplied everything out and got stuck... This is a brilliant solution. One day I will achieve this type of mathematical intuition. Lead the way, blackpenredpen!!!
I think you forgot about the absolute value; Square root of a 2nd power produces absolute value result because both positive and negative values are true.
Dunkoro it is obvious that the number inside the square root is positive，so ignore the absolute value symbol
You could have made it simpler using the (a-b)(a+b)=a^2-b^2 formula.
(x^2+3x)(x^2+3x+2)+1 =
(x^2+3x+1-1)(x^2+3x+1+1)+1 =
(x^2+3x+1)^2-1^2+1 =
(x^2+3x+1)^2
Wow your method and my method are similar .....I had take the whole expression as y and then square it and then assume x to be 500 and multiplied and had taken x^2+3x to be z and at end I got y=z+1 that is y = x^2 + 3x + 1 and it's done
At 3:45, you could also treat the first factor as (x^2+3x+1-1) so together with the second factor you have (x^2+3x+1-1)(x^2+3x+1+1) = difference of squares ((x^2+3x+1)^2 - 1. Plus the extra 1 on the outside you get the perfect square.
The conclusion of this question is : [ First number + second number ^2 ]
the easier way would be to solve it symmetrically:
lets say x equals 501,5 in this case.. then the product would be (x-1,5)(x+1,5)(x-0,5)(x+0,5)+1=
(x^2-2,25)(x^2-0,25)+1=
x^4-2,5*x^2+0,5625+1
which is obviously
(x^2-1,25)^2
I understood everything until the bit at 4:35 - 5:04
What do you mean by "factoring out"?
When a number repeats itself in an addition you can factor it out, basically do the inverse of distributive property. So we have x²+3x+1 repeating in both therms. You can factor it out and you will be left with x²+3x+1 ( x²+3x +1), equivalent to x²+3x+1( x²+3x) + x²+3x+1 (1)
@@AE-rg5rc but thats just squaring it and you don't have two of the sake expression..you don't jave twobx squared plus 3x plus 2 you only,have one
@@leif1075 please Learn some math, this is for grade 6 atleast in asian countries.
@@ViratKohli-jj3wj I know some math, thanks very much..I had a valid question
You may substitute x^2+3x = t and make t(t+2)+1 = t^2+2x+1 so square root of t^2+2x+1 is t+1=x^2+3x+1(same)
Just do the multiplication by hand.
Just say x^2+3x = y
y(y+2) + 1
= y^2+2y +1
=( y+1)^2
Comes out of root as
(Y+1)
= x^2+3x + 1
And put the values
Him: Starts the video with 1+1=2
Me: *ok we are getting somewhere now
Why not substitute y=x^2 + 3x then you can expand that to get y^2 + 2y + 1 = (y+1)^2. Perfect for cancelling out the square root