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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.

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

$b^2+2b$ can never be a perfect square as $b^2+2b+1$ is (except for $b=0$).

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  • $\begingroup$ That is what i thought as well, but can that be presented as valid proof, or do i need to find other way to prove it? $\endgroup$ – Стефан Јовановић Oct 24 '15 at 16:11
  • $\begingroup$ 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. $\endgroup$ – Ian Miller Oct 24 '15 at 16:13

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