Prove that $$ \sum_{i=0}^{k} p^{2i} $$ where $k > 0$ and $p$ is an arbitrary prime, is never a perfect square. I think you can prove it by letting $q = \sum\limits_{i=0}^k a_ip^i$, then expanding $q^2$ and equaling coefficients of $p^l$ in $q^2$ and the original sum, thus showing no such $a_i$ in $q$ exist. But I'm kinda looking for a more elegant solution. Thanks.
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1That looks like a geometric series to me... – TMM May 02 '13 at 23:07
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It is. Its sum is $\frac{p^{2(k+1)}-1}{p^2-1}$. But it doesn't really help me, I'm a newbie at number theory. – user75619 May 02 '13 at 23:15
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Try factoring the numerator. – TMM May 02 '13 at 23:19
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If $k=0$ it is a perfect square so perhaps let $i=1$??? – John Marty May 02 '13 at 23:37
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Let's call it a trivial case and disregard it completely. :) – user75619 May 02 '13 at 23:43
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Ok I agree, and I'll do that. – John Marty May 02 '13 at 23:45
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I've shown that if $p$ is an odd prime $k \equiv 0,-1 \pmod4$, in fact if k is odd it is $\equiv 3 \pmod 8$, but I can't get any further. – John Marty May 03 '13 at 00:26
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2This is related to http://math.stackexchange.com/questions/372367/1x-ldotsxn-perfect-square ,where $x=p^2$ – lsr314 May 03 '13 at 05:21
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@Sophie; in fact, one of the answers to that question answers this as well. – vadim123 May 03 '13 at 23:40
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@Hecke I'm sure the primality of $p$ will allow easier methods here. – Bart Michels Mar 15 '14 at 13:34
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My partial process:
Assume $x^2=\sum_{i=0}^{k} p^{2i}$. Working modulo $p$ gives $x^2\equiv 1\pmod p$ or: $$x\equiv\pm1 \pmod p.$$ Using the formula for power series gives: $x^2=\frac{p^{2(k+1)}-1}{p^2-1}$, or: $$x^2p^2-x^2=p^{2(k+1)}-1\\ (xp)^2+1^2=(p^{k+1})^2+x^2$$ This is in the form $x^2+y^2=z^2+w^2$, which as discussed here, is of the form: $$(xp,1,p^{k+1},x) = (a c + b d , b c - a d , a c - b d , a d + b c),$$ Which lead to some complex identities between $a$, $b$, $c$ and $d$. For example, $bc-ad=1$ implies $(a,b)=(a,c)=(b,d)=(c,d)=1$.
LeeNeverGup
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