I'm looking for Pythagorean triangles with a hypotenuse of length 2021. In other words, let $z=2021$ then, $z^2=x^2+y^2$ for some integers $x$, $y$. Now, when I put a hypotenuse value of 2021 to generate Pythagorean triples using an online calculator, it shows that there are no solutions i.e. no such pythagorean triangle exists.
However, $z^2=2021^2=43^2 \cdot 47^2$ and $43, 47$ are both of the form $4k+3$, so by sum of two squares theorem, since the two exponents are even, $z^2=x^2+y^2$ should be solvable…
I think there might be a flaw in my logic somewhere. Could someone clarify on this and also provide a proof in case such a triangle cannot exist?
Best Answer
Consider the following two articles:
Generating a Triple and Sum of Two Squares.
Since $~43 \equiv 3 \pmod{4},~$ and $~43 ~| ~2021$, the two articles collectively imply that $2021$ can not be the hypotenuse of a primitive pythagorean triple.
That is, if it is a primitive, then you would have to have
$$m^2 + n^2 = 2021$$
which is impossible.
This leaves the issue of whether it can be a non-primitive hypotenuse of a pythagorean triple.
However, since $~43 \times 47 = 2021,~$ this possibility can be manually eliminated by examining the elements in $\{1,4,9,16,25,36\}$.
That is, it is impossible to take any two of those elements and have them sum to either $43$ or $47$.
Therefore, it is game over.
Edit
Upon reflection, both $43$ and $47$ are congruent to $3 \pmod{4},$ so, manual examination of $\{1,4,9,16,25,36\}$ isn't necessary.