I put together a VBA script to find me all of the Pythagorean Triangles from 3,4,5 all the way to 105, 608, 617. Then I arranged them by increasing hypotenuse $c$, and identified all of the triangles where $c$ was a prime number. I then subtracted each $c$ value from the preceeding prime $c$ and found that in every instance the difference between the two values was divisible by 4.

Example 1:

$$12^2+35^2=37^2$$ $$9^2+40^2=41^2$$

and $41-37 \equiv 0 \pmod 4$.

Example 2:

$$9^2+40^2=41^2$$ $$29^2+45^2=53^2$$

and $53-41 \equiv 0 \pmod 4$. And so forth.

Is there a way to prove whether or not this holds true for all prime numbers that can be represented by the square root of the sum of two square integers?


The reason is that the hypotenuse is of the form $$ c = \lambda (u^{2} + v^{2}), $$ where $u, v > 0$ (are coprime and) have different parity, that is, one of them is even, and the other is odd.

If such a $c$ is prime, then $\lambda = 1$, and then $u^{2} + u^{2} \equiv 1 \pmod{4}$.

Hence the difference of two such $c$'s is $\equiv 0 \pmod{4}$.


You can represent every primitive Pythagorean Triple with $\{M^2-N^2,2MN,M^2+N^2\}$, where the last number represents the hypotenuse. For $M^2+N^2$ to be prime, $M^2+N^2\equiv1\bmod4$ or $M^2+N^2\equiv3\bmod4$. The second case can be ignored because $M^2\equiv0\bmod4$ or $M^2\equiv1\bmod4$ for all integers, so $M^2+N^2\equiv0,1,\text{or }2\bmod4$. This means that all prime hypotenuses must be one more than a multiple of four, so their difference is always going to be divisible by four.

  • $\begingroup$ @AndreasCaranti: He means every primitive Pythagorean triple. That is, no common factor. $\endgroup$ – Tito Piezas III Jan 13 '17 at 14:58
  • $\begingroup$ @AndreasCaranti Thank you. $\endgroup$ – AlgorithmsX Jan 13 '17 at 15:00

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