New Zealand Mathematical Olympiad 2019 Question 5
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- เผยแพร่เมื่อ 14 มิ.ย. 2020
- We present a solution to round 1 question 5 from the 2019 New Zealand Mathematical Olympiad. This problem involves describing all values of a quartic expression which is a perfect square.
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He didn't end with its usual quote "it's good place to stop"
:(
lol, I didn't know this was a thing!
:(((
:(
@@MichaelPennMath sorry "this is a good place to stop"... was your usual ending
Meanwhile me in an exam situation:
Well, let's try 1. Nope, doesn't work.
Let's try 2. Yup, got it.
But you gotta find all solutions, so then you would have to prove that there are no more solutions for n >= 3.
@@megauser8512
/- joke-\
/\ you \/
3:26 + 20
I think the easiest way to see that 9n^2-10n+5 is positive for natural numbers is to write it as n(9n-10)+5. The first term is negative only when n=1 where we get a total of 4 .
Or: the vertex of that parabola is at n=10/(2*9)=5/90 for n=1, it has to be >0 for all greater numbers also.
Or just calculate delta 100-4*5*9=-80 neg so the function has the signe of a in this case 9 so positive
I like the first hint you gave out in the beginning of the video. Great job!
8:26 he picks c=5 because c=3 and c=4 have c^2 terms less than 20, which will make the resulting expression negative for low values of n(0,1,2).
ty
Indont think anyone would think of that hint of multiplying by another perfect square..its just so strange and counterintuitive..there must be another way???
@@leif1075 Well , actually we dont like dealing with quartic equations so we want to eleimate the first 2 terms and we observe for a perfect square the 2ab term requires the same ccoefficient as the a^2 term i.e 2*k=k^2 or k is 2 and the coefficients become 4 and 4 which we are then allowed to multiply the the original expression
@@minatisahoo4533 does he mean c equals 5 and NOT c equals 3 at 8:40..because on the board he has written c equals 5 not 3..
@@leif1075 Yes
I did it a slightly different way, completing the square and showing that if there is a solution, it looks like a^2 + b^2 = c^2 and therefore we can use the equation for a Pythagorean triplet. Though my approach is somewhat less elegant!
Please give a detail layout of your solution, thanks!
I inspected the first few numbers and found n=2 worked then did synthetic division to find there was a remainder of 57/( n-2) and figured that meant there were no other solutions.
Synthetic division of what with what?
The multiplication by 4 is a very smart idea. Thank you.
this bounding was also useful on EGMO 6
for n in range(1,100):
for i in range(1,100):
if((pow(n,4) - pow(n,3) + (3 * pow(n,2)) + 5) == pow(i,2)):
print(n)
*python code*
Very well explained!!😃😃😃😃
I did it in a simpler way imo:
The general approach to this kind of problem is to try and bound the expression between two consecutive squares. n^4 seems like a good upper bound because of the -n^3 attached to it, which should overpower all the other terms in the limit. Thus, we shall try to find k such that n≥k implies n^4>n^4-n^3+3n^2+5, or in other words, n^3-3n^2-5>0. The derivative is equal to 3n^2-6n, which is increasing for n≥2, and by inspection n=4 makes the expression positive, so n≥4 should also work. Therefore, k=4.
The next step should be to try to bound the expression below by (n^2-1)^2. Similarly, we shall try to find m such that n≥m implies (n^2-1)^2>n^4-n^3+3n^2+5, ie n^3-5n^2-4>0. Deriving the expression, we get 3n^2-10n, which is increasing for n≥4, and n=6 satisfies the inequality by inspection, so m=6.
Because of that, we can already conclude n≥6 will satisfy the inequality (n^2-1)^2
also love the "tap" on the board @1:06 !!! : )
Can you please feature this nice integral problem? Integrate the (fractional part of 1/x)^k, where k is some natural number, from limits x=0 to 1. The expression for the final answer is to be written in form of a sum, which can be computed for a few values of k.
Ans: sum (zeta(j+1)-1)/(binom(k+j)(k)) from j=1 to \infty
Is it possible to use induction?
Sir we could have looked a perfect square as sum of odd nos and expression suggest that difference of two square is 5
this solution is so hard you can easily solve it with some creative inequalities
edit: as it was pointed out in the comments this proof is wrong I am gonna post a slightly different one in the comments below
first you observe n^3>3n^2 +5 for every n>=4 (1)
so you check n=1,2,3 for solutions and you find n=2 as a solution
and then you make the the argument that n^4 -n^3+ 3n^2+5=4 ]
which leads to 3n^3+4n+4n
proof for (1)
n>=4=>(1/3)*n^2>5=>(3+1/3)*n^2>3n^2+5 and since n>3+1/3=>n^3>(3+1/3)*n^2>3n^2+5
Tres belle remarque bravo
Where are you using the fact that n^4-n3+3n^2+5 has to be a square?
Oh I think I see what you are saying. You are saying n^4-n^3+3n^2+5 is a square less than n^4. But then how can you conclude it is actually less than (n-1)^4. There are 2n+1 squares inbetween (n-1)^4 and n^4?
@@willnewman9783 yes you are absolutely right, instead of (n-1)^4 I should have used (n^2-1)^2 which leads nowhere and makes my proof wrong, Let me try to rectify this
At 11:23 - an alternative is to show that f(n) = 9 n^2 - 10 n + 5 = 0 has no real roots (which is true because 10^2 - 4*5*9 = 100 - 180 = - 80 is not the square of a real number). Since f(n)=0 has no real roots, and f(n) is continuous, it cannot change sign. And since f(0) is obviously 5, f(n) must be positive for all real numbers.
Edited to add: there's also a transcription error at 10:05 or so: A - (2 n^2 - n + 2) = 3 n^2 + 4 n + 16, not 3 n^2 + 14 n + 16. The conclusion is still correct (A is always greater than (2 n^2 - n + 2)^2 for any natural number n).
Why go back to the quartic when have reduced it to a quadratic to find n near the end?
Great job
At 11:40 another easy way to notice it is always greater than 0 is by noticing with n=1 it is equal to 4, and then n=2 is also positive, and then notice than the exponential n^2 is growing at a greater rate than -10n once the n becomes around 3ish etc, so the value of the expression will only become greater, thus always being positive, given n being a natural
At 3:37 hew wrote A = 4n^4-4n^3+12n+5 that 5 should be 20.
He fixed that.
nice, clean handwriting
10:05 3n^2+4n+16. In any case, always greater than 0. Thanks.
Great job sir
What about this :
7n^3-3=m^2
What we can do
For me, the key insight was to try to bound n^4-n^3+3n^2+5 *between* "consecutive" squares, rather than analyze the "squared-ness" of n^4-n^3+3n^2+5 directly. To that end, I started by finding values of a and b such that n^4-n^3+3n^2+5 = (n^2+an+b)^2 cancels as many terms as possible, which yields a = -1/2 and b = 11/8. The net result is that you can strictly bound sqrt(n^4-n^3+3n^2+5) between n^2-n/2+1 and n^2-n/2+2 for large enough n...in fact, for n > 2, so n doesn't even have to be that large. And for even n, n^2-n/2+1 and n^2-n/2+2 *are* integers, fortunately, so you're done. For odd n, there's just one more small step: verify that the unique integer between n^2-n/2+1 and n^2-n/2+2, which is n^2-n/2+3/2, can't square to give you your n^4-n^3+3n^2+5. Wrap it up by checking n=1 and n=2 explicitly.
Hey i have one doubt, will you answer it
@@user-yh6lr8wy9s sure
Brilliant
I have an easier approach for this problem.
From question we let
n^4 - n^3 +3n +5 = k^2 (k is non-negative integer)
Rearrange the equation we get
n^2 ( n^2 - n +3) = k^2 - 5
Note:
1) Both terms at LHS must be non-negative integers, RHS must also be non-negative integers.
2) The term n^2 is a perfect square
3) Factors of 5 are only 1 and 5. Therefore, the gcd of k^2 and 5 can only be either 1 (perfect square) or 5
Case 1 (if k^2 is not a multiple of 5)
n^2 ( n^2 - n +3) = (1)( k^2 - 5)
By comparison
Sub case 1i)
n^2 = 1 --> n = 1
Sub n =1 into following equation
n^2 - n + 3 = k^2 -5
k = sqrt 8 (rejected)
Sub case 1ii)
n^2 - n +3 = 1
No real solution for n.
In fact the LHS term is not a perfect square.
Case 2 (k^2 is a multiple of 5)
We can rewrite the equation as
n^2 ( n^2 - n +3) =5( k^2 /5 - 1)
Where k^2 /5 is still an integer
Since 5 is not a perfect square
n^2 - n + 3 = 5
n = -1 (rejected) or n=2
Sub n=2
n^2 = k^2 /5 -1
4 = k^2 /5 - 1
k = 5
Thus the only solution obtained is when n = 2
Checking the solution by substitute n = 2 into original and u will obtain 25 which is a perfect square.
"easier"
@@particleonazock2246 I just write out my full solution with detail explanation up there...
In short, we dont need to eliminate n^4 and what we have to deal with are just factor pairs of (k^2 -5) and simple quadratic equations
@@steffchoong1518 I can understand your solution, but how do you know when to solve quadratics (or equations in the form of quadatics) with casework? When modular arithmetic will suffice just as easily, that is to say.
Thank you very much for the detail layout of your solution, very good, I understand it, and it is much earsier then Michael Penn's solultion, much appreciated!!
Absolutely fenomenal
Since the problem was to find *all* natural values of n that would make the given quartic expression a perfect square, and it turns out there is only *one* such value, then a "guess" of n = 2 is a very quick solution - which I noticed in less than 5 seconds. Nevertheless, actually *proving* it was very impressive - another great video!
That "it turns out" to have only one solution is basically the entire problem.
I got correct, I started with 1 next got 2 as the solution and gone till 6 and not got any so stopped.
He didn't end this video with his usual quote "it's good place to stop"
There is a much cleaner and easier way to solve this using quadratic residue. n^4 - n^3 + 3n^2 + 5 = z^2. z^2 is comparable to 1 mod 8, and so is the long expression. Hence, there are 4 cases: n odd, z odd; n even, z odd ... Modulo arithmetic immediately reduces it to 2 n odd, z even and vice versa. Then you write n^4 as 4^k(8m+1). After some trivial checking the solution is clear.
Please give a detail layout of your solution, thanks!
Put n is equal to it gives 25 . is first perfect square for that
How did these questions are even made?
Do we need to prove there are no other solutions if you guess a larger c? Or you have proved it?
If you guess a larger c you would have to check more equality equations. If your c was too small the inequality couldn't be proved. So it's in your best interest to guess the smallest c which satisfies.
@@paulstelian97 we are asked to find all n values, how do we know that there are no n values for larger values of c? Pure guessing without proving that c can't be greater than 5 feels like an incomplete solution.
@@Ooopsi The expression is strictly increasing for larger values of c, if the rest of it remains the same. Larger c would just give you more attempts that have Δ < 0, and larger c will give even worse value for Δ.
@@paulstelian97 Alright thanks, I managed to prove that ∆ decreases as c increases when c>5 and that ∆
@@Ooopsi i can't even think of a way to prove it
for Z the solution is -1
Easy: when the sides are equal length.
Can anyone tell me the solution of this functional equation or its source please :)
The problem:
Let α,β ∈ Q* . Find all functions f: R ->R
Such that:
f([x+y]/α)=[f(x)+f(y)]/β
Is Q* only the rational numbers?
I get the Q* now: I think that it means all rational numbers except 0, since if one of those 2 variables is ever 0, then we would be dividing by 0.
Problem:
Let α,β ∈ Q* and find all functions f from
R to R such that f([x+y]/α)=[f(x)+f(y)]/β.
Solution:
Note that for x=y=0: f(0)=f([0+0]/α)=[f(0)+f(0)]/β=2f(0)/β,
so f(0) must be 0, but also for x=t and y=-t:
f(0)=f([t+(-t)]/α)=[f(t)+f(-t)]/β=0, so f(-t)=-f(t) for all
t ∈ R, so f is an odd function. Furthermore, for x=αt and y=0:
f(t)=f(α*t/α)=f([α*t+0]/α)=[f(α*t)+f(0)]/β=f(α*t)/β,
so ^^^f(α*t)=β*f(t)^^^ for all t ∈ R, and also for x=y=α*t:
2*f(α*t)/β=[f(α*t)+f(α*t)]/β=f([α*t+α*t]/α)=f(2*α*t/α)=f(2*t),
so 2*f(α*t)=β*f(2*t) for all t ∈ R, but with ^^^, we have:
2*β*f(t)=2*f(α*t)=β*f(2*t), so 2*f(t)=f(2*t) for all t ∈ R,
but t is just a dummy variable, so we can replace t with x:
2*f(x)=f(2*x) for all x ∈ R, but this implies that if we take
f(x)=its infinite series=sum(n>=1){c_(2n+1)*x^(2*n+1)}, [since
f(x) is odd (see above), all the even terms are 0], we have:
2*sum(n>=0){c_(2*n+1)*x^(2*n+1)}=2*c_1*x+2*c_3*x^3+... =
c_1*(2*x)+c_3*(2*x)^3+...=sum(n>=1){c_(2*n+1)*(2*x)^(2*n+1)},
so 2*c_1*x+2*c_3*x^3+... = 2*c_1*x+8*c_3*x^3+..., but
since 2=/=8, and in general 2=/=2^(2*n+1) for all n>=1, then
2*c_3*x^3=/=8*c_3*x^3 unless c_3=0, and in general
2*c_(2*n+1)*x^(2*n+1)=/=2^(2*n+1)*c_(2*n+1)*x^(2*n+1)
for all n>=1 unless c_(2*n+1)=0, but the 1st term works out,
since 2*c_1*x = 2*c_1*x for all c_1 ∈ R. Since c_1 is just
a dummy variable, then we can replace it by c, so we have:
f(x)=c*x, where c ∈ R, but we must find all c that work with
c * [x+y] / α = f([x+y]/α) = [f(x)+f(y)]/β = [c*x + c*y] / β
c * (x + y) / α = c * (x + y) / β, so either α = β, or c=0.
Therefore, all functions f from R to R that satify
f([x+y]/α) = [f(x)+f(y)]/β, are the following:
f(x) = c*x, with c ∈ R and α=β, OR with c=0 and α=/=β
Great work...keep it up
The solution was not too intuitive. Can anyone suggest some intuitive methods
7:19 I don't understand this argument here. There is no reason to compare the coefficient of the dominating term here because n is not going to infinity.
Sure, but it's also not going to zero, because you're working in the positive integers. So higher order terms will generally tend larger, which is all you need to know for this purpose.
Loving the videos, but your audio is terribly quiet. Perhaps turn up the gain on the mic or video software?
Man.... I really like your videos but I get lost so often, can you suggest me some good videos to start with. I really want to understand everything you do.
In what kind of videos you get lost?
There was no need for checking the first case as A is even and the squared thing was odd ...
Doesn't that depend on the parity of n?
Number 2 will satisfy your fomula end up as 25 and will be a square of 5.
Let n^4-n^3+3n^2+5=m^2 , where m ∊ N …. Eq1.
If m = 1 or 2, then Eq1 has no solutions such that n ∊ N. We can see this as the LHS of Eq1 is ever increasing for n ∊ N, with a minimum value of 8, so m = 1 or 2 are not possible.
So, for the rest of the solution we will assume m ≥ 3. This being the case, we note that 5 - m^2 < 0, where 5 - m^2 ∊ Z.
Eq1 can be rewritten as n^4-n^3+3n^2+(5 -m^2)=0 … Eq3.
Applying Strum’s theorem to Eq3, we find that the equation has two real roots, where one is negative and the other is positive for a given value of 5 -m^2,
where m ≥ 3.
Of course if Sturm's theorem is an over kill for you, then you can just find the critical points of n^4 - n^3 + 3n^2 + 5 and it will be discovered there there is only one, and it occurs when n = 0, and it is a minimum by the second derivative test. This being the case and the realisation that as n goes to infinite or negative infinite, then n^4 - n^3 + 3n^2 + 5 goes to infinite. Now with the minimum at n = 0, then the minimum is at 5 - m^2, which is less than zero and hence there is one negative real root and one positive real root, and so the remaining two roots are imaginary.
Now let k = m^2-5, and place it in Eq3.
⇒n^4-n^3+3n^2-k=0 … Eq4.
Now let’s assume that n = d is a positive real value solution to Eq4, where d ∊ N, and place this in Eq4.
⇒d^4-d^3+3d^2-k=0 … Eq5.
⇒d^2 (d^2-d+3)=k
⇒d^2 divides k. So, let k = td^2
⇒d^2 (d^2-d+3)=td^2
⇒ (d^2-d+3)=t, as d ≠ 0
⇒ d^2-d+3-t=0 … Eq6.
Let’s factorise Eq6 and see if we can find b and d ∊ N that satisfies the equation.
Let (d+b)(d+(3-t))=0
⇒ d^2+d(3-t)+bd+b(3-t)=0
⇒ d^2+d(3-t+b)+b(3-t)=0 … Eq7.
Comparing Eq6 and Eq7, we observe that
-1=3-t+b and 3-t=b(3-t)
⇒ b=1, and hence t= 5.
Placing t= 5 into Eq6
⇒ d^2-d-2 =0
⇒(d+1)(d-2)=0
⇒ d =-1, or d= 2.
Now as d ∊ N, then d= 2 is the only possible solution.
Now with d= 2 and t= 5, and as k = td^2, then k = 20.
As k = m^2-5, then m^2=25, thus m = 5.
Placing m = 5 into Eq3, we find indeed that n = d = 2 is a solution, and it is the only solution for any m as our analysis has shown, as only when t = 5. (I.e. when m^2=25) is an integer solution possible to Eq3.
12:06 Do naturals not include 0? Although if you specialcase n=0 the inequality will still hold.
Some people define them to include 0 and others exclude 0, its quite ambiguous
0 is not a natural
@@chhabisarkar9057 I think 0 IS included in the naturals, and in Romania at least if you want the positive integers (that is, 1,2,3 etc without 0) you say nonzero naturals specifically. Or the notation N*. N includes 0. And when someone says naturals, I assume N and not N*.
@@paulstelian97 Yes, 0 is included in natural numbers set according to ISO. But in olympiad maths forums and communities they prefer to use N as a set of positive integers. Since this notation is ambiguous in many terms, in real contests they prefer to say set of positive integers or rather use Z+ notation. It is important to clarify which notation is used in problem since problem may get extremely obvious (for example, in functional equations substituting zero gives a great information about function itself)
Many North American pedagogies distinguish the "Natural Numbers N" = {1, 2, 3, ...} from the "Whole Numbers W" = {0, 1, 2, 3, ...}.
Definitely confusing, and by no means universal, but worth being aware of.
Haha I remember this last year
3:25 *20 not 5
Sneaky transition to fix it.
Soy el primero! 😅🎉
So, n=-2/5 is a natural number. Which means that this expressions are perfect square. OK!!! 😁😁👍👍
14:38
Please a humble request that make some videos teaching geometry of circumcircle and cyclic quadrilateral and pigeonhole principle a humble request by your Indian fan and subscriber
Prepping for the INMO, I see
Don't worry you won't qualify INMO , RMO , or even PRMO.
How much time is given per question in the Olympiad?
Depends on the phase and the country of the olympiad. Typically in the last phase there are 6 problems split between 2 days. Normaly you have 3 problems for 4.5 hours on each day.
This is the most common format because it is the format used at IMO (International Mathematical Olympiad), so most countries emulate that.
Generaly olympiad questions are really hard, in my country if you do 2 questions out of 6 and manage to get some partial credit in some others (getting some of the steps in the solution, even if you dont complete it), thats often enough to get a bronze medal. If you do 4 questions you are likely to get a silver medal and if you do 5 or 6 you will get a gold medal.
I am not totally convinced by your argument. You have found a solution in the sequence of squares u(n, r) = (2n² - n +r)², but there are 3(4n² -2n + 8) positive integers between your upper and lower limit for 4n⁴ - 4n³ + 12n² + 20, and you haven’t shown that none of these other than the one you found are perfect squares. So in effect you have failed to demonstrate the uniqueness of the solution you found.
The sequence of squares are all the possible squares in between the bounds...... (2n^2-n+2)^2 and (2n^2-n+5)^2... there are only 2 squares in between.
First three terms take n out and equate.
The 16-th Hungary-Israel Binational Mathematical Competition 2005 (imomath.com/othercomp/Hi/Himc05.pdf ) had strange problem number 3 (find all sequences of distinct positive integers x1,x2,...,xn such that 1/2=1/x1^2+1/x^2+...+1/xn^2). There seems to be no full solution nowhere to be found on the internet and all related posts seem to imply it is exceptionally hard, yet it was on the competition (unless its presence on the competition was "trolling" the attendees...). Still I think it's interesting to look at.
Sounds like they were trolling you. I believe this is Project Euler problem 152, which doesn't even require finding all solutions. The problem does list a couple solutions, but I'd hate to see a youtube video spoiling this problem.
@14:38 mood
I tried to solve it from the thumbnail:
I thought a "perfect square" meant a quadratic in the form (an+ b)^2 or a^2.n^2 + 2abn + b^2
And going from there I got n = something crazy
Turns out "perfect square" just meant "a square number"
So... I approached this completely wrong :/
I hated it when you start guessing
Plz do more problems form indian national mathematical olympiad or Regional mathematical olympiad from india 🙏🙏 it's a request
i remember you from math quest
i made a little Python program to check for other solutions. for now, it found a second one: n = 33554535. i checked a couple of times and it looks like it satisfies given conditions. can someone verify it? maybe i made a little mistake somewhere
Wolfram Alpha verifies you made a mistake.
Perfect squares must be congruent to 0 or 1 (mod 4). For the below, mod 4 is used.
Your number is congruent to negative 1. When you plug negative one into the polynomial you get 10, which is congruent to 2. Hence result of the polynomial can’t be a perfect square with your n.
Lol, I competed in that and got that question wrong XD
Here is my solution:
For n ∈ N, when is n^4 - n^3 + 3n^2 + 5 a perfect square?
Since n ∈ N, then n must be >= 1, so let's try n = 1:
n^4 - n^3 + 3n^2 + 5 [at n = 1]
= 1^4 - 1^3 + 3*(1^2) + 5 = 1 - 1 + 3 + 5 = 8,
which is NOT a perfect square, so n = 1 is NOT a solution.
Let's try n = 2, then:
n^4 - n^3 + 3n^2 + 5 [at n = 2] = 2^4 - 2^3 + 3*(2^2)
+ 5 = 16 - 8 + 3*4 + 5 = 13 + 12 = 25 = 5^2,
which IS a perfect square, so n = 2 IS a solution.
Claim: There are NO (integer) solutions for n >= 3.
Proof: f(n) = n^4 - n^3 + 3n^2 + 5 is a perfect square if
and only if (2^2)*f(n) is a perfect square, so 4*f(n) is:
4*f(n) = 4*n^4 - 4*n^3 + 12*n^2 + 20,
so 4*f(n) is between:
4*f(n) - g(n) = (2*n^2 - n + 2)^2
where g(n) = 3*n^2 + 4*n + 16
and
4*f(n) + h(n) = (2*n^2 - n + 4)^2
where h(n) = 5*n^2 - 8*n - 4,
since g(n) and h(n) are both > 0 for all n >= 3,
since all terms in g(n) are positive, so g(3) > 0,
h(3) = 5*(3^2) - 8*3 - 4 = 5*9 - 24 - 4 = 45 - 28 = 17 > 0,
and both g(n) and h(n) increase with increasing n. So,
if there is a solution for n >= 3, then we must have:
4*f(n) + j(n) must = (2*n^2 - n + 3)^2,
where j(n) = n^2 - 6*n - 11 must = 0 for some n >= 3,
but since there are no 2 integers which add to -6 and
multiply to -11, we cannot factor j(n) to a form with
integers, so there are no integer solutions to the
equation j(n) = 0. Therefore, there are no solutions
to the equation f(n) = an (integer) perfect square for
any integer n >= 3. In conclusion, our only solution
to this problem is n = 2.
I started off trying to square (ax^2 + bx + c) and compare it to the original quartic, and got stumped.
I guess I'm not too disappointed in myself for giving up after a couple days, since I would never have thought of this solution. Although maybe I could've come up with a different way. Anyway, great video, I'm always learning and enjoying
That seems like the intuitive way to do it. I'm just a 17 year old kid so maybe not the best person to ask about this haha, but a lot of this solution, while possible to follow, seems very complex and counterintuitive
Now n^4 - n^3 + 3n^2 + 5 = m^2.
So let f(n) = n^4 - n^3 + 3n^2 + (5 - m^2) = 0
Now, you will find that when m > 2, then there are only two real roots to the quadratic for each value of m, where one is positive and the other is negative, and there are two imaginary roots. Note: no solution is possible when m = 1, or m = 2.
So we only need to consider solutions where 5 - m^2 < 0.
You can see what kind of roots we have by differentiating f(n) and finding that f'(n) = 4n^3 - 6n^2 + 6n = n(4n^2 -6n + 6).
Now, when f'(n) = 0, there is only one real number solution, which is n = 0. This turns out to be a local minimum by inspecting f'(n) as n move from -ve to zero to +ve, or by the second derivative test.
Now, at n = 0, then
f(n) = n^4 - n^3 + 3n^2 + (5 - m^2) = 0.
So, f(0) = 5 - m^2 < 0, where m > 2.
This being the case that the only local minimum ( there is no local maximum) at n = 0 is negative and f(n) goes to infinity, as n goes to infinity and when n goes to negative infinity, then we find that f(n) must be of the form
(n + a)(n - b)(n^2 + cn + d), where a and b are +ve (giving us one negative root and one positive root respectively), and n^2 + cn + d is irreducible (i.e. can't be factorised with real numbers).
Try to equate coefficients of (n + a)(n - b)(n^2 + cn + d) to n^4 - n^3 + 3n^2 + (5 - m^2) and after you work through a number of cases, you will conclude that n = 2 is the only solution where m = 5.
If you try it, I will publish the completion of the solution.
3:36-+20, not +5
Good, you fixed it
Great solution. But, the only one anwser is ridiculous.
14:40 😂😂😂mental calculation can sure be a bitch.
15:53 I mean obviously there are no solutions: you imposed that an even number is the square of an odd number.
A is clearly even, but 2n^2 - n + 3 doesn't have to be odd? E.g. n = 1 gives an even number; am I missing something?
@@stefanenache9002 you're right, in fact one can show with modular arithmetic that 2n^2-n+3=n(2n-1)+3 is odd if n is even and even if n is odd.
Anyone who can prove or disprove this ....
If x^p = y^p (mod n ) then x=y (mod n)...
Its converse is true...
Thanks..
The first statement is false.
Example:
Let n = 8 and let x=1, y=3 and p=2 then x^2=1, y^2=9 == 1 mod 8
So x^2 = y^2 mod 8 but x != y mod 8
Proving the converse:
Assuming p is an integer and x == y mod n
We can use induction here for a bit of overkill because really the proof showing x^2 == y^2 immediately shows this holds for natural powers
for p = 1 this is clearly true, assume the statement holds for all k
n^4 - n^3 + 3n^2 + 5 = k^2
n^4 - n^3 + 2n^2 - n^2 + 5 = k^2
n^2(n^2 - n + 2) - n^2 + 5 = k^2
n^2(n - 2)(n + 1) - n^2 + 5 = k^2
This suggests solutions of -1, 0, and 2. The first two are not natural numbers, 2 is *a* solution.
But I don't know how to prove there aren't more solutions.
I did a similar mistake but note that 2n^2 + n^2 (not minus) and in the last parentesis the factorization is wrong
he has been working out lol
Dude's a mountain climber
what is that deep breathe
Lol
Thank you
Why c=5?