Given: $$a + b - c = 2$$ $$a^2 + b^2 + c^2 - ab + ac + bc = 1009$$
Solve for a positive integer triplet $(a, b, c)$.
I'm not exactly sure how to approach this problem. The second equation doesn't seem factorable. I've tried plugging in different values of $a, b, c$ found in the first equation into the second equation, but that hasn't been very helpful. Any suggestions would be appreciated! Thanks!
Substitute $c=a+b-2$ in the second equation, and you get $$ a^2 + a b + b^2 - 2 a - 2 b = 335 $$ and thus $$ \frac{3}{4} (a+b)^2 + \frac{1}{4} (a-b)^2 - 2 (a+b) = 335 $$ or $$ (3a + 3b - 4)^2 + 3 (a-b)^2 = 4036 $$ There are finitely many integer solutions to $x^2 + 3 y^2 = 4036$...