Pythagorean bliss

Number Theory Level 4

Let $p$ be a prime number of the form $p = 4n + 1$ ( $n \in \mathbb N$ ).

How many different right triangles exist with integer lengths for their legs, such that the hypotenuse is $p$ ?

That is, how many unordered pairs of positive integers $\{a, b\}$ exist such that $a^2 + b^2 = p^2$ ?

Example

If $p = 17$ , there is only one pair: $\{8, 15\}$ , because the 8-15-17 triangle is the only kind of right triangle with integer sides and hypotenuse equal to 17.

On the other hand, for $p = 65$ we have two options: a 16-63-65 triangle and a 33-56-65 triangle. However, this does not count because 65 is not a prime number.

It may be one, but can also be more It must be one It can be 0, 1, or more It may be one, but can also be zero

1 solution

Arjen Vreugdenhil
May 8, 2016

Fact 1 : Every primitive Pythagorean triple $a^2 + b^2 = c^2$ has the form $a = u^2 - v^2,\ \ b = 2uv,\ \ c = u^2 + v^2$ (after swapping $a$ and $b$ if necessary).

Fact 2 : Every prime number of the form $p = 4n + 1$ can be written uniquely as the sum of two non-zero perfect squares.

Combine these:

$p = 4n+1 = u^2 + v^2$ uniquely with $u > v$ ;
let $a = u^2 - v^2$ and $b = 2uv$ ;
then $a^2 + b^2 = p^2$ is the unique Pythagorean triple with the desired property.

Thus there is always a unique solution; the answer is $\boxed{\text{one}}$ .

For instance, if $p = 157$ then $u = 11$ and $v = 4$ because $11^2 + 6^2 = 157$ ; so we take $a = 11^2 - 6^2 = 85$ and $b = 2\cdot 11\cdot 6 = 132$ ; then $a^2 + b^2 = 85^2 + 132^2 = 24\:649 = 157^2$ .

Why every 4n+1 prime is equal to the sum of squares

Mr Yovan - 5 years, 1 month ago

That is a rather profound result in number theory, known as Fermat's Two-Square Theorem .

Arjen Vreugdenhil - 5 years, 1 month ago

Arjen ..the link given in comments doesn't talk of Uniqueness ???

Piyushkumar Palan - 3 years, 10 months ago

Suppose that $p = a^2 + b^2$ and also $p = c^2 + d^2$ , with $a > c$ even and $b < d$ odd.

Let $d = (a+c)/2, e = (a-c)/2, f = (d+b)/2, g = (d-b)/2$ . It is easy to check that $de = fg$ .

Let $u = \gcd(d,f)$ and $v = \gcd(e,g)$ , and let $x = d/u = e/v$ , $y = e/v = d/u$ . Then $p = a^2 + b^2 = (d + e)^2 + (d - e)^2 = (ux + vy)^2 + (uy - vx)^2 = (u^2 + v^2)(x^2 + y^2),$ showing that $p$ is not a prime number.

Example: Take $p = 65 = 8^2 + 1^2 = 4^2 + 7^2$ . Then we get

$d = 6$ , $e = 2$ , $f = 4$ , $g = 3$ ;
$u = 2$ , $v = 1$ , $x = 3$ , $y = 2$ ;
$p = (2^2 + 1^2)(3^2 + 2^2) = 5\cdot 13$ .

Arjen Vreugdenhil - 3 years, 10 months ago

Pythagorean bliss

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