Let $ S = \frac{b^n-1}{b-1} = b^{n-1}+b^{n-2}+...+b+1$ Since $ 2\mid b$, $ S\equiv b^2+b+1 (mod 8)$ Since S is a perfect square and it's odd, $ S\equiv 1 (mod 8)$. So $ b^2+b+1 \equiv 1 (mod 8)$, i.e. $ 8\mid b(b+1)$. As $ (b,b+1) = 1$, $ 8\mid b$ or $ 8\mid (b+1)$, but since b is even, the latter is impossible, thus b is divisible by 8.

STARS wrote: Let $ b$ be an even positive integer for which there exists a natural number n such that $ n>1$ and $ \frac{b^n-1}{b-1}$ is a perfect square. Prove that $ b$ is divisible by 8. Let $b=2k$ for some natural number $k$ Now as $n>1$ so let's check for $n=2$ Then $2k+1=x^2$ now as $x^2 =(0, 1,4)mod 8$ for every natural number $x$ then $2k+1=x^2$ is only possible if $8$ divides $2k$ hence for $n=2$ $8$ will divide $b$ Now for every $n\geq 3$ 8 divides $(2k)^n$ Then $\frac{(2k)^n - 1}{2k-1} = x^2$ then $\frac{(2k)^n - 1+x^2}{2x^2}=k$ now clearly $(2k)^n - 1+x^2$ must be even it is only possible when $x^2=1mod 8$ and as $n\geq 3$ then 8 divides $(2k)^n$ hence at least $4$ divides $k$ then as $b=2k$ so $8$ divides $b$ $\blacksquare$

what is Estonian Math Comp?

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parmenides51 wrote: ItzARubix wrote: what is Estonian Math Comp? there is always google http://www.math.olympiaadid.ut.ee/eng/html/index.php Oh TIA