# At what base is square root of 120 natural number

I got this problem i need to solve: At what base b, where b>2 , (120)b equals x2, where x is in decimal number system? I need to find all bases b, and i need to see the process of finding answer, so i can do it myself.

• Hint: $(120)_b=b^2+2b=(b+1)^2-1$. – Apple Oct 24 '15 at 16:05

$(120)_b = b^2+2b$
$b^2+2b$ can never be a perfect square as $b^2+2b+1$ is (except for $b=0$).
• Yes. Or include Apple's hint and express it as $(b+1)^2-1$ to show that it is always one less than a perfect square. – Ian Miller Oct 24 '15 at 16:13