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Is it possible to construct a right triangle with integer side lengths and a hypotenuse of $2^{100}$?

After looking at a list of pythagorean triples, I couldn't find a hypotenuse of a right triangle with integer side lengths that was a power of $2$. So I would be inclined to think the answer to this question is no. I was thinking of using a modular arithmetic argument to show that $a^2 + b^2 \neq 2^{200}$ for any positive integers $a,b$, but I couldn't find a mod that would work (I tried $3,4,5,6,7$).

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  • $\begingroup$ If the hypotenuse is length $2^{100}$ then the equation should be $a^2+b^2=(2^{100})^2=2^{200}$ $\endgroup$
    – ASKASK
    Commented Jan 30, 2016 at 3:47
  • $\begingroup$ @ASKASK : You can code $2^{100}$ in MathJax as 2^{100}. $\qquad$ $\endgroup$
    – Michael Hardy
    Commented Jan 30, 2016 at 3:48
  • $\begingroup$ @MichaelHardy bad, I'm on mobile $\endgroup$
    – ASKASK
    Commented Jan 30, 2016 at 3:48

1 Answer 1

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The hypotenuse is either the largest of a primitive Pythagorean triple, or a multiple of one.

The largest number in a primitive triple is always odd.

Does $2^{100}$ have any odd factors?

EDIT: because I was asked, adding a quick proof that of primitive triple $(a,b,c)$ with $c > \max(a,b)$, $c$ is always odd.

Consider $a^2 + b^2 = c^2$.

First observe that both $a$ and $b$ can't both be even, because both sides would be even, and since they have a common factor, this is not a primitive triple.

So either $a$ and $b$ are both odd or they have different parities.

Suppose the former. Then the $LHS$ would be $2 \pmod 4$, while the $RHS$ would be $0 \pmod 4$, giving a contradiction.

So the latter has to be the case. The sum of an even and an odd number is always odd, giving the required proof.

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  • $\begingroup$ Why is it always odd? $\endgroup$
    – user19405892
    Commented Jan 30, 2016 at 4:11
  • $\begingroup$ @user19405892 Added a quick proof. $\endgroup$
    – Deepak
    Commented Jan 30, 2016 at 4:23
  • $\begingroup$ Why does a multiple of the largest of a pythagorean triple have to be odd? You seem to be claiming that $2^{100}$ must be the hypotenuse in a primitive pythagorean triple. $\endgroup$
    – user19405892
    Commented Jan 30, 2016 at 4:26
  • $\begingroup$ I never claimed that a multiple of the largest of a PPT has to be odd - it can very well be even - nor that $2^{100}$ necessarily has to be the largest of a PPT. The hypotenuse of a right triangle with integer sides is always either the largest of a PPT or an integer multiple of the largest of a PPT. $2^{100}$ is clearly even, so it can't be the largest of a PPT. That leaves the second case to be tested. For the second case to hold, at least one factor of $2^{100}$ has to be odd (namely the largest of the PPT under consideration). I asked you if $2^{100}$ has any odd factors at all? $\endgroup$
    – Deepak
    Commented Jan 30, 2016 at 5:03
  • $\begingroup$ (except $1$ of course, that should be obvious and trivial). $\endgroup$
    – Deepak
    Commented Jan 30, 2016 at 5:09

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