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Prove that $x^2 + y^2 = z^n$ has a solution in $\mathbb{N}$, where $n$ belongs to $\mathbb{N}$, the set of all natural numbers.

My Attempt:
Without loss of generality we can assign $z$ a perfect square value.

For $n=1 , x^2 + y^2 = z = k^2$ clearly has many solutions in natural number( since x,z,k are Pythagorean triplets).

Let the expression have a solution for $n = m$. Let it be $( x_1, y_1, z_1 = k_1^2 )$. Therefore,
$ x_1^2 + y_1^2 = z_1^m = k_1^{2m} $

For $n = m+1$, \begin{align*} x^2 + y^2 &= z^{m+1}\\ \Rightarrow x^2 + y^2 &= z^mz \end{align*} clearly $(k_1.x_1, k_1.y_1, k_1.z_1 = k_1^2 )$ is a sol. of the equation, i.e.,
$(k_1.x_1)^2 + (k_1.y_1)^2 = z_1^m k_1^2$
$=>(k_1.x_1)^2 + (k_1.y_1)^2 = z1^2$
Thus, it is proved by induction.

Can You please tell me if my method is right. Is it justifiable to take z a perfect square?

Also, can you provide a non-induction proof? ( but before you do please tell me if my method is correct or not)

user354595
  • 189

2 Answers2

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The short answer is, your method is correct.

A non-inductive version using the same basic idea would be to simply note that

$$(3\cdot5^{n-1})^2+(4\cdot5^{n-1})^2=(3^2+4^2)5^{2(n-1)}=5^2\cdot5^{2(n-1)}=(5^2)^n$$

so $(x,y,z)=(3\cdot5^{n-1},4\cdot5^{n-1},5^2)$ is a solution in positive integers to $x^2+y^2=z^n$ if $n\ge1$.

If you want to get fancier and show that there are "primitive" solutions -- that is, solutions with $x$, $y$, and $z$ relative prime, then one approach would be to start with

$$5=1^2+2^2=(1+2i)(1-2i)$$

from which you can get

$$5^n=(1+2i)^n(1-2i)^n=(a_n+b_ni)(a_n-b_ni)=a_n^2+b_n^2$$

The trick here is to show (inductively) that if $5$ doesn't divide $a_nb_n$ then it doesn't divide $a_{n+1}b_{n+1}$ either.

Barry Cipra
  • 79,832
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We know that the set of sums of two squares is closed by multiplication

$$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2\qquad(*)$$ It follows that if we take $z=a^2+b^2$ with $a$ and $b$ arbitrary integers we have $$z^n=(a^2+b^2)^n=X^2+Y^2$$ after apply iteratively the precedent operation $(*)$ so $X$ and $Y$ are required solutions. We see this way there are infinitely many solutions.

Piquito
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