# Prove that $\sum\limits_{i=0}^{k} p^{2i}$ ($p$ is prime) is never a perfect square

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.

• That looks like a geometric series to me...
– TMM
May 2 '13 at 23:07
• 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. May 2 '13 at 23:15
• Try factoring the numerator.
– TMM
May 2 '13 at 23:19
• If $k=0$ it is a perfect square so perhaps let $i=1$??? May 2 '13 at 23:37
• This is related to math.stackexchange.com/questions/372367/… ,where $x=p^2$ May 3 '13 at 5:21

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 $(a,b)=(a,c)=(b,d)=(c,d)=1$.