Prove that $x^2+y^2=z^n$ has a solution $(x, y, z)$ in $\mathbb{N}$ for all $n\in\mathbb{N}$
I tried to prove this by induction, but couldn't. ( This was probably because the solution for some $n$ isn't necessarily related to the solution for $n+1$)
I can't seem to see any other way other than induction for proving the statement. Any help/hints on solving this problem, or any alternative approach will be highly appreciated :)
On
(No induction necessary) Start with a known pythagorean triplet $(x,y,z),$ for example $(3,4,5).$ If you multiply all three terms in the equation $x^2+y^2=z^2$ by $z^{2n-2},$ the left hand side will still be a sum of squares and the right hand side will be the $n$-th power of the square of the original $z.$
One explicit solution would be $(3.5^{n-1},4.5^{n-1},25)$
On
There is a neat trick to this problem that uses complex numbers. Note that $|a+bi|^2=a^2+b^2$. So if $m=a^2+b^2=|a+bi|^2$ and $n=c^2+d^2=|c+di|^2$, then $mn=|(a+bi)(c+di)|^2=|(ac-bd)+i(ad+bc)|^2$. This tells us that the numbers that are sums of two squares are closed under multiplication.
Now, let us start with any pair of numbers, $a,b$. Then $a^2+b^2=n$, or $|a+bi|^2=n$. Taking powers, we have $|(a+bi)^k|^2=n^k$.
Its easy to find a solution for $n=2$, $3^2+4^2=5^2$. Its also easy to find a solution for $n=3$, $2^2+2^2=2^3$.
Now notice that if $x^2+y^2=z^n$ then $$(x\cdot z)^2+(y\cdot z)^2=x^2\cdot z^2+y^2\cdot z^2=z^2(x^2+y^2)=z^{n+2}$$
Hence by induction it is possible for all $n$.
Edit: Just noticed I skipped the $n=1$ case. $1^2+1^2=2$. Done.