Existence of a Pythagorean Triple with one side given.

Question

I am curious about the answer to the question:

Does there exists a pythagorean triple with $n$ as one of the sides for all $n\geq 3$ ?.

Your answers and comments will mean a lot.

Answer 1

We have

$n^2 + (\frac{n^2-1}{2})^2 = (\frac{n^2+1}{2})^2$.

When $n$ is odd and greater than or equal to $3$, the second and third terms are both positive integers.

When $n$ is even and not a power of $2$, we can write $n=2^kn'$ where $n'$ is odd and greater than or equal to $3$. Take a solution for $n'$ and then multiply through by $2^k$.

Finally, for any $k\geq 2$ we have

$(2^k)^2 + (2^{2k-2}-1)^2 = (2^{2k-2}+1)^2$.,

where the second and third terms are positive integers by our assumption on $k$. This covers all powers of $2$ greater than or equal to $4$.

Answer 2

If $n$ is odd, we have $$n^2 + \left(\dfrac{n^2-1}{2}\right)^2 = \left(\dfrac{n^2+1}{2}\right)^2$$ If $n$ is even we have $$n^2 + \left( \dfrac{n^2}{4}-1 \right)^2 = \left( \dfrac{n^2}{4} + 1 \right)^2$$