Can I ever be natural?
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The interesting thing is that 7 is the value of infinite nested root √(42+(√42+√(42+√(...)))). Actually, if you replace 42 with x*(x-1) for x>1, the limit will be exactly 'x'.
Excellent observation! Making 42 a really special number 😊
@@JeanYvesBouguet Not so special. Take any √(m+√(m+x)) and do m = x² - x
For instance x=5, √(m+√(m+5)) take m = 25 - 5 ...
@@Neodynium.the_permanent_magnet 42 is certainly a special number though
@@Neodynium.the_permanent_magnet I wonder if there is a name for numbers that fit this pattern. I.E. some pattern n(x), where n(7)=42, n(5)=20,...
@@Rougarou99 It works for any N
Case 2 can be even more quickly ruled out through parity. (4n² + 29 is odd, (2n+2)² is even)
Exactly
There is much simpler way to find n from 4n^2 + 29 - is square
Lets 4n^2 + 29 = k^2
then
(k-2n)(k+2n) = 29 | 29 is a prime number so 29 = 1*29
then we got system of linear equations:
k = 2n + 1 => k = 15
k = 29 - 2n => n = 7
The rest goes the same
This is good
Yes it's a shame not to use the fact that 29 is prime, given the opportunity
Ikr
Yeah, that's how I did it.
For any odd prime p, the only natural numbers whose squares differ by p are (p ± 1)/2.
Guess I'll delete my comment now....
Another way: let sqrt(m+7)=a and sqrt(m+a)=b, then m+7=a^2 and b^2-a=m, replacing m from the latter to the former eq we have: b^2-a+7=a^2, after multiplying by 4 and adding 1 to both side, we can rearrange the last eq into (2a+1)^2-(2b)^2 = 29. Hence (2a+2b+1)(2a-2b+1)=29 which is prime therefore it can only be factored into 1 and 29. 2a+2b+1 is the bigger factor thus 2a+2b+1=29 and 2a-2b+1=1, solving for a and b we get a=b=7 thus m=b^2 - a =42.
You wrote the the difference of squares wrong, it should be (2a-2b+1)(2a+2b+1) as you wrote later.
thanks for noticing the obvious typo. Just fixed it.
@@galveston8929 why would you or anyone think to multiply by 4 at all though or multiply by anything even?
@@leif1075 from a^2 + a = b^2 - 7, you can complete the square in terms of a and obtain (a+1/2)^2=b^2+29/4, which motivates the multiplication by 4 to obtain a difference of integer squares.
@@leif1075 that's part of completing the squre. When you do this many times, then you'll do it by heart.
42 the answer to life the universe and everything !!!
Surely it's simpler if you notice that if 4n^2 + 29 is a perfect square, p^2 say, then p^2 - (2n)^2 = 29 and therefore (p - 2n)(p + 2n) = 29. So as p + 2n is larger of the two factors we must have p + 2n = 29 and p - 2n = 1. Thus p = 15 and n = 7.
While there is nothing wrong with what you have written, and indeed (at least as far as I'm aware) it is a simpler solution, you should probably have specified that your last line is only possible because 29 is a prime (and hence irreducible).
Case 3: We have 4n^2+29=(2n+3)^2 so 4n^2+29=4n^2+12n+9 so 12n=20 which does not provide another solution.
Hey folks - I edited this video. Let me know if you think the sound is better since I tried to clean it up.
Oh so that’s why it wasn’t live at the usual time 😂
Nah. This was edited a couple days ago lol
I don’t control when it goes up i just edit lol
There is a bit of a thing everytime he says "s", but other than that sounds great
@@roryisatall1 thank you for that feedback. I can try on future videos to fix that.
05:30 we can assume (2n)²+29=k² for natural k.
So we get 29=(k+2n)(k-2n)
Which as k and n are positive integers, and 29 is prime, there is only one possibility,
k+2n=29
K-2n=1
so we get n=7.
The 3rd case for n doesn't work out, because
4n²+29=(2n+3)²
4n²+29=4n²+12n+9
As always, the 4n² cancels out, so we get that
12n=20
n=20/12=5/3 which isn't an integer, so it doesn't count here
Came here with the same thought and glad to see someone else had come to the same conclusion
Easier just to check mod 3.
@@Grassmpl oh really? I'm not that familiar with modulo stuff, could you show me?
@@AnAverageItalian you can substract 3 on the right equation to get 4n^2 +29 = (2n)^2 (mod 3), then eliminate the n we have 29 = 0 (mod 3) which is simply wrong
@@afa12345 subtract*
My solution: sqrt(m + 7) should be integer, so m = k*k - 7, then all we need is k*k - 7 + k to be a square. We can check all k < 7 and for all k > 7 this cannot be square since k * k < k*k - 7 + k < (k + 1)*(k + 1). So k = 7 is only solution and m = 42
12:58
In an age of specialization, your have procured a micro-niche. "The human time stamp of a signature phrase"
Another solution : Let m and q be such that m+7=q²
(1) m + √(m+7) = q² - 7 + q = q² + 2q + 1 - (q+8) and hence m +√(m+7) < (q+1)²
(2) Also, if q > 7 then m > 42 hence m + √(m+7) > m + 7 = q².
From (1) and (2) we get that if q >7 then q² < m + √(m+7) < (q+1)² and hence √(m + √(m+7)) cannot be an integer.
After that, all that is left is to check the values of q from 1 to 7.
Title sounds like you’re having an existential crisis
For 4:50 onwards, 4n^2 is already a perfect square, and the consecutive squares are always some odd number away, so 4n^2 and 4n^2 + 29 could be 196 and 225 (the largest possible answers), giving n=7; or smaller squares separated by an (odd) number of these "steps". But since 29 is prime, there are no others. (Then find m as you did.)
I don't think 29 being prime gives the result you think it does. The 'step' between squares wouldn't be 29 divided by the step size. Instead, for a single step difference, you look for the smaller square to be (29-1^2)/(2*1)=7. For the next odd step, it would be (29-3^2)/(2*3), then (29-5^2)/(2*5), etc.
@@jursamaj My reasoning is that if they are 3 odd numbers away, then call them k-2, k, and k+2, so the difference is 3k, which cannot be since 29 is prime. Same for any other odd number.
@@guidomartinez5099 Exactly right. If p is prime, p = h^2 - k^2 = (h + k)(h - k) gives h + k = p & h - k = 1 as the only possible factors.
guys idk what these symbols are. why do we need to find men?
Once again it would seem that one can find a solution, m=42 by inspection but the difficulty lies in proving it's the only solution.
Yes, I saw that solution immediately by thinking "what if the inner sqrt is also 7" but it's not obvious that's the only solution.
But that also wasn't the question in the video title
Finally found the question to the answer to life, the universe, and everything!
What a fun problem and solution! I couldn't help modifying the problem and giving it a go myself!
Or, let y=sqrt(m+7). Then m=y^2 -7. Then the original expression converts to sqrt(y^2-7+y). Therefore y^2+y-7 must be a perfect square. But this is always less than (y+1)^2=y^2+2y+1, and for y>7, it is greater than y^2 (therefore in between 2 consecutive squares and not a perfect square). So we only have to check seven values of y in y^2+y-7 to see if any is a perfect square, and get y=7. We know m=y^2-7, so m=42.
found m=42 by noticing that: m+7 must be a perfect square, otherwise m+sqrt(m+7) can't be a perfect square. Thus, m must be divisible by 7 and adding 7 to it has to be a perfect square too, which leaves 6*7=42 as the only possible solution. This gives me m=42 and n=7 as the only possible solution.
You need m + sqrt(m+7) to be a square, so in particular it has to be a natural number, so m+7 is a square. Call it m+7 = s^2 so m = s^2 - 7 and the goal is for s^2 + s - 7 to be a square. This is between (s+1)^2 and s^2 unless s is small; in particular either s^2 + s - 7 = s^2 or s^2 + s - 7
7:27 it's much easier to move the square term to the right and factor because after you factor the difference of square it has to be even [ (2n+2-2n)(2n+2+2n) = 2(4n+2) ] and you have an odd number on the left. Same applies to the (2n+3) case; (2n+3-2n)(2n+3+2n) = 3(4n+3) which is divisible by 3 but 29 is not divisible by 3. Also slightly easier for (2n+1) because you don't have to multiply it out as one of the factor becomes one when you do the difference of squares: (2n+1-2n)(2n+1+2n) = 1(4n+1) = 29 so n = 7
Super easy to follow explanations, thank you Michael
Just came here to see if my solution is right I solved it in 20 secs just by THUMBNAIL 💀
Proud to be an Aisan
Have you just discovered the ultimate question of life, the universe and everything? We've had the answer for a while but have been struggling to work out what the question was.
Nice reference to the hitchhiker's guide to the galaxy saga.
If you say that m = n^2-7 (which is must be), the algebra becomes much simpler.
set sqrt(m+7)=n and m=n^2-7 then sqrt(m+sqrt(m+7))=sqrt(n^2+n-7) between n and n+1
my intuition at the start told me that it's unlikely for there to be 2 perfect squares that fit into the radicals in the presented form, so my guess was that the inner root must be the same as the outer root and hence sqrt(m+7)=7 which directly gives the solution m=42. of course it's not a prove that there are no other solutions, but it felt unlikely to me
The actual answer to the title is: “Dude what are you talking about, i is always imaginary”.
Exactly what I thought-of course; it's only logical.
@@lewsouth1539 LLAP, other Nimoy.
haha 42 “pretty popular number on the internet”
thanks for explaining why the "+" case goes wrong
If we replace 7 with an arbitrary x, then m = x^2 - x.
RJ
42, always a good 'guess' I guess......
I read the title and really thought this was going to be about something other than math.
You can find the 42 solution pretty quickly, by making the substitution sqrt(m+7)=7. That makes the outer radical equal to the inner radical, and since the inner is a perfect square from that equation, the outer is also. From there it is easy to solve for m=42
Yes. All the math is only needed to prove there is never another solution. It's pretty trivial to guess that m = 42 is a solution.
We now have the question to the answer...
Can someone please explain to me how he got 4n^2+16n+16 from 4n^2+32? at 5:45
That part is a little odd, I agree.
What he's trying to do is find a perfect square that's just slightly bigger than 4n² + 29 to be the upper bound. That square is (4n² + 4)² or 4n²+16n+16. The +32 value is just an intermediate step which establishes the inequality (assuming n>=1, which it is).
It's confusing because he works through it in the "equality" order, so the +16 value seems to just appear out of nowhere. It makes more sense if you start with that and go backwards.
If x = √(m+√(m+c)), then is it always true that natural x = c?
Update: there's some recursion. Can we use:
if x = √(m+x) then
x = √(m+√(m+x))
if we knew that the initial condition in the first sentence was always true, then the problem can be reduced to: 7 = √(m+7)
Where then m is easy
Привет, я сейчас в 11 классе и мы считаем интегралы в уме(не шутка). Очень интересное видео, спасибо. Вы решали сборник задач Демидовича по матанализу?
This problem inspired me to try another problem: given a natural number c, find natural numbers m and n such that root(m + root(m + c)) = n. Following the same basic method that Michael showed us, I worked out a really quick way to do it: find the odd factors of (4c+1); these will be called "k values". For each value of k, calculate (4c+1-k^2)/4k; each natural number that results will be a possible "n value" (note that if 4k is greater than half the value of 4c+1-k^2, the result cannot be natural). For each possible n value you can get up to 2 possible m values, which are calculated using m = 1/2×(2n^2+1±root(4n^2+4c+1)). In the case where you end up with more than 1 ordered pair, you have to check them manually.
I really liked the discussion on the extraneous solutions, and that 3rd case shouldn't give you a solution.
Find m for y:
y = √(m+√(m+7))
Let x be natural where:
x = √(m+x)
Substitute right side into itself:
x = √(m+√(m+x))
If we let x be 7 we have the equation:
7 = √(m+√(m+7))
Depending on the existence of natural:
7 = √(m+7)
49 = m + 7
m = 42
Verify y is natural when m = 42:
y = √(m+√(m+7))
y = √(42+√(42+7))
y = √(42+7)
y = 7
A solution for m = 42
As another approach... We manage to get that we need 4m^2 + 29 to be a perfect square. Therefore, we have some k such that 4m^2 + 29 = k^2. Let p = k - 2m. Therefore, k = 2m + p. So we have 4m^2 + 29 = (2m + p)^2. Expanding the RHS, we get 4m^2 + 29 = 4m^2 + 4mp + p^2. This means that 29 = 4mp + p^2. But the right hand side is clearly divisible by p. So, 29 has to be divisible by p. That means that p is either -29, -1, 1 or 29. Using an argument similar to what Michael did, we can eliminate -29, -1 and 29. Therefore, the only candidate is p = 1, so we have 4m^2 + 29 = (2m + 1)^2. Solve as Michael did.
You can also prove it simpler:
Let b = sqrt(m+7) => m = b^2-7 and for sqrt(m+sqrt(m+7)) to be natural we need sqrt(m+7) to be natural.
So if m is a solution to our problem then b = sqrt(m+7) is natural and we have sqrt(b^2-7+b) is a natural number in other words b^2+b-7 is a square.
But for all b > 7 we know that b^2 < b^2+b-7 < b^2+2b+1 = (b+1)^2. So that we must have b 2 since b = sqrt(m+7).
So we check for b = 3,4,5,6,7 and find that b^2+b-7 is only a square if b = 7.
Lastly we have to check if m = b^2-7 = 42 is really a solution of our problem and indeed sqrt(42+sqrt(42+7)) =7.
Thus m = 42 is the only solution.
Let u= sqrt(m+7). u must be a natural because m is a natural number too. Then m=u^2-7. So the problem will turn into sqrt(u^2+u-7) = x, where x belongs to N.
Then x^2=u^2+u-7. But this implies that:
x^2-u^2=u-7 or (x+u)(x-u)=u-7
If x>= u, then u >=7, but x+u>u-7 , So equality must be satisfied and u=7=> m=42.
If x
Perhaps less elegant, but a convenient process I went through:
We know that if (m+7) is not a perfect square, then the whole equation cannot be a natural number. Hence, the possibility space for m is restricted to m = q^2-7 for nonnegative integers q. We can exclude q=0,1,2 because those values would make sqrt(m+7) negative and thus make n imaginary. The first handful of valid m values are 2, 9, 18, 29, 42, 57, 74, which we can use to show both that a solution exists, and that that solution is unique. These are associated with the q values 3, 4, 5, 6, 7, 8, 9.
We get the following results from these options, simplifying the sqrt(m+7) parts to just the associated q value.
sqrt(2+3) = sqrt(5), invalid (4 shy of 3^2=9)
sqrt(9+4) = sqrt(13), invalid (3 shy of 4^2=16)
sqrt(18+5) = sqrt(23), invalid (2 shy of 5^2=25)
sqrt(29+6) = sqrt(35), invalid (1 shy of 6^2=36)
sqrt(42+7) = sqrt(49) = 7, valid!
sqrt(57+8) = sqrt(65), invalid (1 above 8^2=64)
sqrt(74+9) = sqrt(83), invalid (2 above 9^2=81)
Each time, you add 1 to the difference between m+sqrt(m+7) and q^2. This gap grows linearly. However, in order for there to be at least two solutions, there would need to be _quadratic_ growth in that gap. As a result, 7 is the only q value that works, and thus m=42 is the only valid integer solution.
I just sat the thumbnail and immediately recognized the common sub expression
a=sqrt(m+7) and b=sqrt(m+a), and hence assumed a=b and thus m+7=49, just like in the power tower n=x^x^x^...^n.
To be more strict, and don't assume anything, just expand further
a^2=m+b => a^2-b=m
b^2=m+7 => b^2-7=m
Then match both
a^2-b=b^2-7
And the symmetry is evident, hence a=b=7
Sqrt (m+7) must be rational. If not m+sqrt(m+7) is irrational and therefore non squared, so m+7=n^2.
n^2+n-7=k^2 -> n^2+n-(7+k^2)=0. We can finish from here just like the video or using the DOQ=Prime argument.
That's a lot of case checking, and I hate case checking if I don't have to.
Let n = √(m + √(m+7)), and note that only the square root of a perfect square can be a natural.
If m is a natural, then for m+√(m+7) to be natural, we must have √(m+7) is natural, so (m+7) is a perfect square, call it a^2 (where a ∈ ℕ). So m = a^2 - 7. Substitute for m:
n = √(m + √(m+7)) = √(a^2-7 + a). Square both sides:
n^2 = a^2 + a - 7. Now rearrange to a quadratic in a:
a^2 + a - 7 - n^2 = 0
a = (-1 ± √(1 + 28 + 4n^2))/2 = (√(4n^2 + 29) - 1) / 2. For a to be natural, 4n^2+29 must be a perfect square, call it b^2 (where b ∈ ℕ). So a = (b-1)/2.
So b^2 - 4n^2 = 29 or (b+2n)(b-2n) = 29. That has one solution among the naturals, where (b+2n) = 29 and (b-2n) = 1, since 29 has only one positive factorisation, and (b+2n) > (b-2n).
That sole solution yields b = 15 and n = 7. Hence a = (b - 1)/2 = 7 and therefore m = a^2 - 7 = 49 - 7 = 42.
Check: √(42 + √(42+7)) = √(42 + √49) = √(42 + 7) = √49 = 7 = n, as calculated. There are no more solutions and no cases to check.
My solution:
square both sides gets: n^2 = m + sqrt(m+c) (1). Set b=sqrt(m+c) (2)
solve (2) for m gets b^2 - c = m. (3)
Substitute (2) and (3) into (1) gets n^2 = b^2 - c + b.
Two solutions for n to be natural:
First b=c (4) gets n^2 = c^2. Substituting (4) into (3) gets m=c^2 - c.
Second b=-c-1 (5) gets n^2 = c^2. Substituting (5) into (3) gets m=c^2 + c + 1.
Thus at c = 7, m=42 or 57.
Check sqrt(42+7) = +/- 7 => 42 +7 = 49 or 42 - 7 = 35. 49 is a perfect square so n is a natural number.
sqrt(57+7) = +/- 8 => 57 + 8 = 65, 57-8 = 49. 49 is a perfect square so n is a natural number.
Edit, I guess that was supposed to only be a principle root so 42 is the only solution since b must be positive. So the full solution for m is m = {c^2 - c, if c>=0; c^2 + c + 1, if c < 0}
If you generalize it "backwards" - for any Natural n such that root(m + root(m + c)) = n, there are infinitely many integer pairs (m,c) that can satisfy the equation, and for each value of m, c=(n^2-1)^2-m. And (0,n^4) is always a solution, as well. The maximum value of m is n^2. When c=7, there is precisely one m (42) producing one n (7), as demonstrated - but for, say, c=17, there are two pairs that produce different n values: (19,17]) = 5 and (272,17) = 17. c=8 is the lowest positive integer with multiple m and n solutions.
11.33 "...42. A pretty popular number on the internet..." Just beautiful.
I came up with 42 as a valid solution just by thinking about what was under the radicals. If m+7 was a perfect square, then if I can make the inner radical, sqrt(m+7), equal to "7", then the outer radical would be the same thing -- sqrt(m+"7"). Therefore, m+7 should be 49, which makes m equal to 42.
just by looking at it, i just figured out that sqrt(m+7) could be literally the same as sqrt(m+sqrt(m+7)) and therefore sqrt(m+7) = 7
sometimes you might have some weird math intuition, but it's better to always check stuff regardless
You can also simply notice 29 is odd, so it is the sum of 2 consective numbers, which means it is thr difference of their squares, namely 14 and 15. From their, since 14=7x2, its square id 4×7^2, which lends 7.
We can get m = 42 pretty much by inspection. After all, if m + 7 is a perfect square, then sqrt( m + 7 ) = k and we ask which k makes m + k a perfect square, which we already know 7 is an answer. To make k = 7 we just need m + 7 = 7^2, or m = 42. Of course, this doesn't prove it's the only solution!
Surely it's basic number theory that the sum of successive gnomons between two perfect squares is divisible by the number of gnomons being summed!
It's just an arithmetic series of consecutive odd numbers and:
case 1) If of an odd number of terms, then equal to the middle term times the number of terms; and
case 2) if of an even number, say 2k, of terms, then equal to the 4k/2 times average term and again divisible by 2k.
Thus 29 being prime, the only perfect squares that it can separate are 14^2 and 15^2.
MPenn is just David Lynch of mathematics; the only distinction is MPenn gives solutions...
42 was easier to find by taking m+7=n^2 and m+n=q^2, then equating n=7=q to satisfy both equations. Of course that says nothing about finding all solutions.
I just thought: what if we get sqrt (m+7) to equal 7, that way you know the outside sqrt will also be 7. Therefore, 7^2 - 7 = 42. Answer is 42.
the answer seems so obvious once you find it out. it could work for any number. if the problem was sqrt(m + sqrt(m + 6)) then it would be 30 since 6^2 - 6 is 30
5:31 "Rewriting" something as something entirely different just because the inequality will still hold is probably the most confusing trick in the entirety of algebra.
I’m getting lazy. I would of put that equation in to an excell spread sheet and run a bunch of numbers. Yes , I have an engineering degree. Fell free to joke about me.
At the start you squared the quantity n^2-m. After that you should have kept in mind that m
Cool. Nice trick with the bounding inequalities.
i saw the video title and ran a python script checking every number. can you imagine my face when i saw the answer? of all the numbers in the universe...
well basically you square the 7 from the equation and substract it and get 42
I cant prove it, but it seems for natural numbers A, where A > 0,the only natural number solution to sqrt(m+sqrt(m+A)) is A(A-1). So it generalizes real nicely I think
See my reply above.
I was just like: "If square root of m + 7 is 7, then we have the same problem again and an easy solution". I just looked at the thumbnail, if all values are asked it is naturally harder
Case 3: 4n^2+29=(2n+3)^2
=4n^2+12n+9
20=12n
n is not natural, so no solution there.
sure a nice way to get the day started, talking for myself here. I got a friend to hop on discord and we watched it together first thing in the morning. Keep up the good work!
Both case 2 and case 3 have solutions if m is allowed to be a rational number and that √(m+√(m+7)) is also allowed to be rational
11:04 . There should not be any self checking if you were cautious enough to see that n^2 - m >=0 before squaring at 1:27.
I already guessed 42 at the start, because sqrt(42+7)=7, which repeats!
sqrt(n^2+n-7)=n to get the only solution n=7 m=7^2-7=42
I'm sure this video goes into something interesting and important, but 42 jumped at me as an obvious hypothesis.
I just guessed x=42 by looking at the inner radical and seeing what can make it a square.
of course it's that number
the answer to life and the universe and everything
I just wrote down every square number, its root and itself minus his root and then you see that there is an interception at and only at 7 with m equals 42.
nice job michael (who needs capital letters anyway?)
Shouldn't it be 2n^2 instead of -2n^2 in de quadratic equation?
Plugging a^2-7 into m could be much faster.
I immediately said 42 from the top of my head, but i didn't know how to prove it
For the 3rd case, you get n=5/3 which does not give a solution.
You seem to really like brute forcing problems.
I find this problem very fascinating.
I observed that the answer can be found very simply, actually
If √(m + √(m + 7)) has to be a natural number, and m also has to be natural, then (m + 7) has to be equal to 7^2. √(m + 7) has to result in 7 so that the following √(m + 7) can be a perfect square.
Thus, 7^2 - 7 = m, which also happens to give us our neat little 42 answer. :D
Very true 42 is a very popular number on the internet.
m=42
M=m^m=1.5013094*10^68
You can be whatever you want if you set your mind to it
Obvious solution - sqrt(m+7) = 7
42 is a popular number? why?
m = k^2-7 -> k^2+k-7 = a^2 -> (2k+1+2a)(2k+1-2a) = 29 -> k = 7, m =42.
I thought about it for a second and got m=42
Me instantly trying 42 out and going yessssss
Why didn't you consider the case of 2n+3
If you wanna be natural, become a log.
anyone else solved this like in 30 seconds?
m = n^2-7 (n >0)
m+sqrt(m+7) = n^2+n-7 = (n+k)^2
If 3 = 7, k >= 0
n = (k^2+7)/(1-2k)
k = 0 because n > 0
n = 7, m = 42
(Is there an error in this proof?)
Probably we can generalize that if sqrt(m + sqrt(m + n)) is natural, then m = n^2 - n.
Interesting idea. Care to prove it? (I think you’re correct btw)
@@NotoriousSRG It's quite simple. Just plug them in and simplify.
sqrt(n^2 - n + sqrt(n^2 - n + n))
= sqrt(n^2 - n + sqrt(n^2))
= sqrt(n^2 - n + n)
= sqrt(n^2)
= n
@@ultimatewierdness n could be anything though. How does this guarantee it’s a natural number? The claim is m=n^2-n and that m is natural. However, if n=π, then sqrt(π^2-π+sqrt(π ^2-π+π) thus sqrt(π^2-π+π)=sqrt(π^2)=π. However π is transcendental.
@@NotoriousSRG If m is natural and m = n^2 - n then n has to be natural as well (or negative, I suppose).
Regardless, all this is is a convoluted way of writing any number, as the simplification showed. Of course you can plug any n into it because it simplifies to n (that might however mean m is not natural).
It is pretty easy to show that m=n^2 -n is always a solution, but there can be others. For example, n = 8 and m = 1 is a solution.