## Algebra 1

$a=3$
We can call the consecutive integers $a$, $a+1$ and $a+2$. Now using the pythagorean theorem, $a^2+(a+1)^2=(a+2)^2$ Expanding this, we get $2a^2+2a+1=a^2+4a+4$. When we simplify, we get $a^2-2a-3=0$. $(a-3)(a+1)=0$ so $a$ must equal $3$