Question

Three consecutive odd integers are such that the square of the third integer is 33 less than the sum of the squares of the first two. One solution is −5​, −3​, and−1. Find three other consecutive odd integers that also satisfy the given conditions.

Answer 1

If $x$ is the first integer, then

$(x+4)^2=x^2+(x+2)^2-33\iff x^2-4x-45=(x+5)(x-9)=0$

The other possibility is then $x=9$, so the other two integers would be $x+2=11$ and $x+4=13$.