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Prove that $\forall n \in \mathbb{N},$ there exist integers $x,y,z$ such that $x^2+y^2=z^n$.

I know that this is an induction proof, and I am assuming this needs to be broken up into cases where n is even and where n is odd.

But how do I even start doing this?

Austin Mohr
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  • Do you consider zero an integer? – 5xum Jul 26 '17 at 13:39
  • Yes. But using Induction, am I allowed to define 4 variables like that? – 1011011010010100011 Jul 26 '17 at 13:39
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    Then you don't need induction, since you can always just set $x=y=z=0$ and the proof is over. – 5xum Jul 26 '17 at 13:40
  • Well, lets say we can't do that.. How would we prove this? – 1011011010010100011 Jul 26 '17 at 13:41
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    prove it's true for some base case n, prove that if it holds for some n it holds for n+1, the rest follows. –  Jul 26 '17 at 13:42
  • @RoddyMacPhee, 'prove that if it holds for some n it holds for n+1'. That part doesn't seem trivial at all to me... – Evargalo Jul 26 '17 at 13:45
  • if it holds for n then multiply both sides by z we get $$z\cdot x^2+z\cdot y^2=z^{n+1}$$ when will this work. in two consecutive cases implying the second form the first if it does it works for all values of n greater than that lower bound. edit: it works when z is a square if we can find one sum that works for it. –  Jul 26 '17 at 13:47

2 Answers2

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Let's assume you mean "nonzero integers", otherwise the claim is obviously true (since $x=y=z=0$ satisfies the equation).

I would advise you to split the statement into two statements:

Statement 1:

If $n$ is even, then there exist integers $x,y,z$ such that $x^2+y^2=z^n$.

Statement 2:

If $n$ is odd, then there exist integers $x,y,z$ such that $x^2+y^2=z^n$.

Proof for statement $1$:

The statement is equivalent to

For all $k\in\mathbb N$, there exist integers $x,y,z$ such that $x^2+y^2=z^{2k}$.

which can be proven by induction.

For $k=1$, the statement should be easy to prove.

For $k\to k+1$, look at what happens to your equation when you multiply it by $z^2$.

With statement $2$, again, you can rewrite it to

For all $k\in\mathbb N$, there exist integers $x,y,z$ such that $x^2+y^2=z^{2k - 1}$.

which is even simpler to prove for $k=1$, while the $k\to k+1$ step is practically identical to the previous statement.

5xum
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  • unless they consider 0 a natural number, then 0^0 is undefined. –  Jul 26 '17 at 13:45
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    0^0=1 si well defined for integers. You just have to not extrapolating to calculate some limits of functions. – Evargalo Jul 26 '17 at 13:47
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    @RoddyMacPhee $0^0$ doesn't appear anywhere in the statement. However it is true that there does not exsist a triple of nonzero integers $x,z,y$ such that $x^2+y^2=z^0$, you are correct. – 5xum Jul 26 '17 at 13:47
  • 0^0 is undefined: 0^anything = 0 and anything ^0 =1 doesn't combine. https://www.youtube.com/watch?v=BRRolKTlF6Q –  Jul 26 '17 at 13:54
  • @RoddyMacPhee Yes, but this is irrelevant to my answer because $0^0$ does not appear anywhere in the statement OP is trying to prove (even if you do consider $0$ a natural number. – 5xum Jul 26 '17 at 13:55
  • Any chance you can elaborate where you write 'the statement should be easy to prove'. How can I just say that $x^2+y^2=z^2$? What is that, Pythagorean Theorem? Or do we actually say that $x=y=z=1$. – 1011011010010100011 Jul 26 '17 at 14:13
  • @sgerbhctim The statement is not "$x^2+y^2=z^2$. The statement is "There exist such $x,y,z$ that $x^2+y^2=z^2$". And a statement like that is proven by finding some $x,y,z$ that satisfy the equation. That shouldn't be hard. – 5xum Jul 26 '17 at 14:15
  • Gotcha. Much appreciated. – 1011011010010100011 Jul 26 '17 at 14:17
  • @Roddy this answer discusses why $0^0$ is often defined to be one. In other words in some contexts it makes sense to say it is not defined, but in a number of other contexts it is defined to be one. – Χpẘ Jul 26 '17 at 16:03
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It suffices to give explicit solutions. I'll give them using $z=5$.

$n=1$: $\quad 5=1^2+2^2$

$n \to n+1:$ $\quad 5^n = x^2+y^2 \implies 5^{n+1} = (x-2y)^2+(2x+y)^2$

This follows from the Brahmagupta–Fibonacci identity.

This identity has a nice interpretation with complex numbers, which in this case is: $$ 5 = |1+2i|^2, \quad 5^n=|x+yi|^2 \\\implies 5^{n+1} =|x+yi|^2 |1+2i|^2 = |(x+yi)(1+2i)|^2 = |(x-2y)+(2x+y)i|^2 $$

lhf
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