list of "Pythagorean triple" equations for $a^2+b^2=2c^2$

Question

I have come up with an equation recently from a project that I'm doing. it's like the Pythagorean theorem except it's $2c^2$ rather than the familiar $c^2$, then I decided that I wanted a list of all the equations that could give me all "Pythagorean Triples", but I don't know where to start (except guess and check).

If anyone is willing, I would like some help on creating this list. I have already found a few, but I can't see any pattern that links them together besides c looks like its always a multiple of a prime number that is one more than a multiple of four.

• $x^2+x^2 = 2x^2$
• $x^2+(7x)^2=2(5x)^2$
• $(7x)^2+(17x)^2=2(13x)^2$
• $(7x)^2+(23x)^2=2(17x)^2$
• $x^2+(41x)^2=2(29x)^2$

Answer 1

If you know the formula for regular Pythagorean triples, you can easily adapt it to your case. Given $p^2+q^2=c^2$ render

$2c^2=2(p^2+q^2)=(p+q)^2+(p-q)^2\equiv a^2+b^2$.

So $c$ is just the hypoteneuse of an ordinary Pythagorean triple and your $a$ and $b$ are just the sum and difference of the corresponding legs. For example, a hypoteneuse of $5$ corresponds to legs of $4$ and $3$, whose sum and difference are $7$ and $1$, so you render $7^2+1^2=2×5^2$.

Answer 2

I don't think that you can find a list of all equations, because of how the equation works. though I have found a method of calculating the amount of solutions for a given $c$ assuming $a$,$b$, and $c$ are natural numbers and are not equal to the same value.

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1. choose $c$,
2. factor into prime numbers,
3. ignore 2 and all primes one less than a multiple of 4,
4. put the coefficients of the remaining primes in the equation $\frac{(2x_1+1)(2x_2+1)(2x_3+1)...-1}{2}$

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ex: $c=3900$
$3900=2^2*3^1*5^2*13^1$
ignore $2,3$
coefficient of $5^2$ and $13^1$
the answer is $\frac{(2*2+1)(2*1+1)-1}{2}=7$