Question:
Verify that3, 4, 5 is the only primitive Pythagorean triple involving consecutive positive integers.
Suppose $\displaystyle a^2+b^2=c^2$ are integers. Without loss of generality, $\displaystyle 0<a<b<c$.
$\displaystyle x^2+(x+1)^2=(x+2)^2$
$\displaystyle 2x^2+2x+1=x^2+4x+4$
$\displaystyle x^2-2x-3=0$
$\displaystyle (x-3)(x+1)=0$
$\displaystyle x=3$ or $\displaystyle x=-1$
But we have already supposed all values are positive, so we discard $\displaystyle (-1,0,1) $ and are left only with $\displaystyle (3,4,5)$.