It's Integral right triangle

Number Theory Level 1

Suppose $p$ is a prime such that $p≡3(\mod 4)$ . Then how many non-congruent right angle triangle with integer sides exists such that one of it's side is $p$ ?

Enter $-1$ if you think there are infinitely many such triangles

The answer is 1.

1 solution

A Former Brilliant Member
Jun 13, 2020

Let the lengths of sides of the right triangle be $p, x, y$ with $x>y$ . We have to consider two cases :

(i) $p$ is the hypotenuse.

In this case, $p$ can be expressed as $a^2+b^2$ where $a$ and $b$ are coprime integers. For both $a$ and $b$ odd, $p\equiv 2 (\mod 4)$ . For one of them even and the other odd, $p\equiv 1 (\mod 4)$ . For both of them even, $p\equiv 0 (\mod 4)$ .

Hence there is no solution in this case.

(ii) $p$ is one of the legs :

In this case $x^2-y^2=p^2\implies (x+y)(x-y) =p^2$ . Since $p$ is prime, $p^2$ has three factors , viz. $1,p,p^2$ .

So the only possibility is $x+y=p^2, x-y=1\implies x=\dfrac{p^2+1}{2},y=\dfrac{p^2-1}{2}$ .

Thus, in all there is only one such integer sided right triangle

Case (i) is more complex than this. We must have $p^2 = a^2 + b^2$ (not just $p$ ), and $p^2$ is congruent to $1$ modulo $4$ . However, since $p \equiv 3 \pmod{4}$ , $p$ is an irreducible element of the Euclidean domain $\mathbb{Z}[i]$ of Gaussian integers, and since $p^2 = (a + ib)(a - ib)$ , we deduce that $p$ must divide either $a + ib$ or $a - ib$ , which means that $a = pA$ and $b = pB$ where $A^2 + B^2 = 1$ , and hence one of $a,b$ must be $0$ . Thus no triangle is possible in this case.

Mark Hennings - 12 months ago

Any pythagorean triples is of the form $(m^2+n^2,2mn,m^2-n^2); m \in Z , n \in Z$ here $m^2+n^2$ is the hypotenuse, $\therefore$ hypotenuse of any integral right angle triangle must be able to be represented as the sum of squares of $2$ integers which is not in the case $\because p≡3 (\mod 4)$ and it is a prime see Fermat's theorem on sum of two squares. . No need to include complex solution.

Zakir Husain - 12 months ago

Well, that all depends on what was meant by $a$ and $b$ in the first proof... At the least, it needs to be more clearly expressed.

Mark Hennings - 12 months ago

Nope. According to my choice (I have mentioned that the sides of the triangle are of lengths $p, x, y$ ), $p^2=x^2+y^2$ , and not $a^2+b^2$ .

A Former Brilliant Member - 12 months ago

It's Integral right triangle

The answer is 1.

1 solution

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