Find the number of right triangles that exist such that each side length is a positive perfect square.

Question

anonymous · Answer

@Directrix

anonymous · Answer

@terenzreignz @Callisto

terenzreignz · Answer

Well, suppose we have such a triangle :)
Then its dimensions could be neatly represented as $\large a^2$ , $\large b^2$  and $\large c^2$  where a, b, and c are integers :)

terenzreignz · Answer

Without loss of generality, let's have $\large c^2$  be the hypotenuse

terenzreignz · Answer

That can only mean, by the Pythagorean theorem, that
$$\Large (a^2)^2 + (b^2)^2 = (c^2)^2$$
$$\Large a^4  +b^4 = c^4$$

Where a, b, and c are integers...

terenzreignz · Answer

Something is amiss, don't you think? ^_^

directrix · Answer

number of right triangles --> My conjecture is zero exist.

terenzreignz · Answer

Conjecture? Their existence kind of contradicts Ol' Fermat :D

anonymous · Answer

Okay, because there are sums of squares that are squares themselves, and there are sums of cubes that are cubes themselves, but nothing higher, right?

terenzreignz · Answer

sums of cubes that are cubes themselves? Not of two cubes.... three cubes, but never 2.

case in point: Fermat's Last Theorem

anonymous · Answer

OK, great, thanks.

terenzreignz · Answer

:)