Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$.
My attempt:
$(a+1)(a-1)+(b+1)(b-1)=c^2+1$
This form didn't help so I thought of $\mod 3$, but that didn't help either. Please help. Thank you.
diophantine equationselementary-number-theory
Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$.
My attempt:
$(a+1)(a-1)+(b+1)(b-1)=c^2+1$
This form didn't help so I thought of $\mod 3$, but that didn't help either. Please help. Thank you.
Best Answer
If $(a,b,c)$ is a solution to $a^2 + b^2 - c^2 = K,$ where your case is $K=3,$ then you get another solution with $$ (a+2b+2c, 2a+b+2c, 2a+2b+3c). $$ Starting with $a,b,c>0,$ you can apply this again and again, the entries grow, forever.
See http://en.wikipedia.org/wiki/Tree_of_primitive_Pythagorean_triples
The matrix calculation that shows that the value of my letter $K$ does not matter is $$ \left( \begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right) $$ In the matrices with the 2's and 3's, usually the one on the left is the transpose of the one on the right in the multiplication shown, but this time they are symmetric so the transpose is the same. Anyway, for any $K,$ you just need to find one solution $(a,b,c)$ with $a,b,c \geq 0$ and, for preference, $\gcd(a,b,c) = 1,$ and then this process gives infinitely many positive solutions for the same $K,$ and they stay primitive as well (gcd 1).
This can be written as a single equation which is fairly readable in this case, $$ \color{blue}{ (a+2b+2c)^2 + (2a+b+2c)^2 - ( 2a+2b+3c)^2 \; = \; a^2 + b^2 - c^2}. $$