Find two pairs of relatively prime positive integers $(a,c)$ so that $a^2+5929=c^2$. Can you find additional pairs with $gcd(a,c)>1$?

Question

This question was asked before, but I was wondering if there's a different approach for this problem.

Find two pairs of relatively prime positive integers $(a,c)$ so that $a^2+5929=c^2$. Can you find additional pairs with $gcd(a,c)>1$?

So, if I was given c instead, and tried to find a and b, then that would be easy. We can write 2c as a sum of squares provided that c is a product of primes congruent to 1 mod 4, then we can find s,t by using the identity and be able to get a and b. I'm trying to use the same approach here but with no luck.

$b = (s^2 - t^2)/2$ 5929 = 77^2

Not sure if it would have helped in finding a and c, but 77 = 7*11, which is not congruent to 1 mod 4, so I cannot write this as a sum of two squares and I cannot apply the identity here.

I'm not really sure how to apply the same techniques here. Any help would be appreciated.

Answer 1

Pythagorean primitive triplet constructions are always very useful. It says that if $a^2+b^2=c^2$, then if $(a,c)$ are relatively prime, $a=2uv$, $b=u^2-v^2$ and $c=u^2+v^2$ for some $u,v$. Hence it is required to write $b$ as a difference of squares. What better than $39^2-38^2=77$? (this is a standard trick, namely that $(n+1)^2-n^2=(2n+1)$, with $n=38$). Now, $2uv=2964$ and $u^2+v^2=2965$, and we have that $2964^2+77^2=2965^2$, where of course $2964$ and $2965$ are co-prime.

To give another answer, we have that $(c-a)(c+a)=7^2 \cdot 11^2$. So find a pair of numbers summing to $121$ and whose difference is $49$. These will turn out to be $85$ and $36$, and their difference is $5929$ as well. These numbers are co-prime too.

If you are looking for non co-prime pairs, nobody stops you from scaling! For example, take any triplet containing $7$, say $7^2+24^2=25^2$. Multiply both sides by $11^2$ to get $77^2 + 264^2=275^2$. Then $264$ and $275$ are not coprime, they are both multiples of $11$.

To give just one more example, $11^2 + 60^2=61^2$, and multiplying by $7^2$, we have that $77^2+420^2=427^2$, and $420$ and $427$ are not co prime because both are multiples of $7$.

Answer 2

For a primitive Pythagorean triple we have:

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

$a=2mn, b=m^2-n^2, c=m^2+n^2$

$m, n$ relatively prime and not both odd.

Then 77, being odd, cannot be $2mn$, so has to be $m^2-n^2$. This equals $(m+n)(m-n)$, and must be $11·7$ or $77·1$. You can find $m$ and $n$ from either of those factorizations and work from there.

To get solutions with a common factor greater than 1, use the above method with 7 or 11 as the odd "leg" of the "right triangle", then multiply the squares by $11^2$ or $7^2$ to pump the leg up to 77.