$a^2+b^2=c^2$ with certain conditions

Question

$a^2+b^2=c^2$ for $a$ and $c$ being a prime numbers and $b$ being a number that is a factor of at most 4 not necessarily distinct prime numbers

My progress. From the fact that both sides of the equation must be odd $b$ is an even number. Say $b=2b_1$. With this substitution and by subtraction of $b^2$ from both sides we get $a^2=4b_1-c^2$ or $a^2=(2b_1-c)(2b_1+c)$ As $a$ is a prime number and expressions in parenthesis are different form each other we have $2b_1-c = 1$ and $a^2=2b_1+c$ This is where I stopped. I' ve tried to use some modular arithmetic or play around and get some more equations by I couldn't. What should I do now? Pls help, I'd prefer to get a hint firstly.

Answer 1

You should get $a^2=c^2-b^2=(c-b)(c+b)$. If $a$ is prime, $c-b$ must be $1$ and you can substitute in to find the triple is $(a,\frac 12(a^2-1),\frac12(a^2+1))$. Pick $a$ an odd prime, compute $b,c$ and check their factors. If $a=5$ you get the $(5,12,13)$ triangle, which is acceptable. If $a=7$ you get $(7,24,25)$ which is not.

Answer 2

I developed a formula that I think addresses your criteria. It generates the subset of all Pythagorean triples where $GCD(A,B,C)$ is an odd square which includes all primitives. The formula $$ A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2$$ generates the table

$$ \begin{array}{c|c|c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 & k=5 & k=6 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 & 13,84,85 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 & 45,108,117 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 & 85,132,157 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 & 133,156,205 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 & 189,180,261 \\ \hline \end{array} $$ Any set can contain $C=P_C$ where $P$ is prime but only $Set_1$ (where $n-1$) can contain $A=P_A$ because $$A=(2n-1)^2+2(2n-1)k\quad=\quad(2n-1)(2n-1+2k)$$ and we can see that if $n>1$, then $A$ is composite.

In this sample we can see double primes in $(3,4,5), (5,12,13), (7,24,25), (11,60,61)$ and the formula for finding such is $ n=1\implies A=2k+1\quad B = 2 k (k + 1)\quad C=2 k^2 + 2 k + 1$.

As for your other criteria, $B$ is always a multiple of $4$. This means for a maximum of $4$ primes

$4\times 1\times 1 :(3,4,5), \quad4\times 1\times 3:(5,12,13), \quad4\times 2\times 3:(7,24,25), \quad4\times 3\times 5:(11,60,61)$

but not such as $\quad 4\times 3\times 3\times 5:(19,180,181) $ because there are too many primes. If you can find values of $k$ where both $2k+1$ and $2 k^2 + 2 k + 1$ are prime and where the $n$ of $4n$ has, at most $2$ primes, you have your solution. One approach would be to try all of $n=p$ where $p$ is prime but you still have to veryify the "primality" of $A$ and $C$.

Good luck.