Prove that for any integer $n$ $\geqslant$ $3$ there is a Pythagorean triangle with one of its legs having length n. For which integers n will there be a primitive Pythagorean triangle with n as one of its legs.

For the first part I tried to start induction with $n = 3$. If we take the other legs to be 4 and 5 then $3^2 + 4^2 = 5^2$

But I'm not sure how to proceed.


For natural $u,v$ the numbers $$a=u^2-v^2$$ $$b=2uv$$ $$c=u^2+v^2$$ satisfy $$a^2+b^2=c^2$$

If the given number $N$ is even choose $u=\frac{N}{2}$ , $v=1$

If $N$ is odd , choose $u=\frac{N+1}{2}$ , $v=\frac{N-1}{2}$

This way you will even get a primitive pythagorean triple containing $N$

  • $\begingroup$ Thanks that helps. $\endgroup$ – user442253 Sep 7 '17 at 22:18

Break the problem into two cases.

Spose $n=2k, k>1$. Let $u=\frac{n}{2}+1$ and $v=\frac{n}{2}-1$. Obviously $u,v\in\mathbb{N}$.

Next, write $$a=u^{2}-v^{2}=2n,$$ $$b=2uv$$ and, $$c=u^{2}+v^{2}=\frac{n^{2}}{2}+2.$$ It is easy to check that no matter what numbers $u,$ and $v$ are, $$a^{2}+b^{2}=c^{2}.$$ Now, $a^{2}=4n^{2}$ and $b^{2}=4(uv)^{2}$, and since $n$ is even, $c$ is even. Therefore $c^{2}=4M^{2}$ for some $M\in\mathbb{N}$, and $$4M^{2}=c^{2}=a^{2}+b^{2}=4(n^{2}+(uv)^{2}).$$ This proves the result for even $n>2.$

For the odd case $n=2k+1$ with $k>0$, let $a, b,$ and $c$ be written in terms of $u,$ and $v$ as above, but choose $u=\frac{n+1}{2}$, and $v=\frac{n-1}{2}.$ Then $a=n,$ and the result follows from the identity $a^{2}+b^{2}=c^{2}.$


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