sinankaral53 wrote: $ a - )$ Find all prime $ p$ such that $ \dfrac{7^{p - 1} - 1}{p}$ is a perfect square $ b - )$ Find all prime $ p$ such that $ \dfrac{11^{p - 1} - 1}{p}$ is a perfect square here it is the Turkish solution

Attachments:
soru_2.pdf (67kb)

Can someone translate solution ?

Here my solution

Attachments:
Solution.pdf (24kb)

It's harder to read in Turkish than to solve this problem!

Since $gcd(q^{\frac{p-1}{2}}-1,q^{\frac{p-1}{2}}+1)=2$ for any odd integer $q$, if $px^2=q^{p-1}-1$ for an integer $x$ and an odd prime $p$, than either Case1:$q^{\frac{p-1}{2}}-1=2py^2$ and $q^{\frac{p-1}{2}}+1=2z^2$ or Case2: $q^{\frac{p-1}{2}}-2=2y^2$ and $q^{\frac{p-1}{2}}+1=2pz^2$ for some integer $y$ and $z$. A)In Case1: $6|7^{\frac{p-1}{2}-1=2py^2}$ ->$3|py^2$. When $p=3$ we have $\frac{7^2-1}{3}=4^2$. If $p\geq 2$ then $3|y^2$ ->$9|7^{\frac{p-1}{2}-1}$ -> $3|\frac{p-1}{2}$. Let $k=\frac{p-1}{6}$. then $2z^2=7^{3k}+1=(7^k+1)(7^{2k}-7^k+1)$ implies that $7^{2k}-7^k+1$ is a perfect square. But is not possible as $(7^k-1)^2\geq7^{2k}-7^k+1\geq(7^k)^2$. In Case2:Since 2 is a quadratic residue $mod7$, but $-1$ is not We conclude that $p=3$ is only prime for which $\frac{7^{p-1}-1}{p}$ is a perfect square. B)This time 2 is not a quadratic residue $mod11$ and we have only Case 2 to consider. $p|11^{\frac{p-1}{2}}+1$ thus $2y^2=11^{\frac{p-1}{2}}-1=-2(modp)$ -> $p|y^2+1$ -> $4|p-1$.Therefore $11^{\frac{p-1}{2}}-1=2y^2$ implies $11^{\frac{p-1}{4}}+1=u^2$ which is impossible as $u-1$ and $ u+1$ cannot be both powers of 11; or $11^{\frac{p-1}{4}}+1=2u^2$ which is impossible 2 is not a sqyare residue $mod11$.So there in no prime $p$ for which $\frac{11^{p-1}-1}{p}$ is a perfect square.

Why do we have 3|py^2 ?????

Domination1998 wrote: Why do we have 3|py^2 ????? Is that too obviously ? $6|7^{ \frac{p-1}{2}}-1=2py^2$ so $3|py^2$.

i wanna know why $4|(p-1)$

Domination1998 wrote: i wanna know why $4|(p-1)$ Lemma: If $p=4k+3$ is a prime and $p|a^2+b^2$ then $p|a,p|b$.

ok thank you $ 11^{\frac{p-1}{4}}+1=u^2 $ but why we can have this?????

any explanation?????

can i know w sir???? why we must have that true???

It has been reached that $11^{\frac{p-1}{2}}-1=2y^2$, with the conclusion $11^{\frac{p-1}{4}}+1=u^2$, or else $11^{\frac{p-1}{4}}+1=2u^2$. Denote $11^{\frac{p-1}{4}} = x$; then it has been reached that $(x-1)(x+1) = x^2-1=2y^2$. Since $\gcd(x-1,x+1) = 2$, it means that either $x+1 = u^2$ and $x-1=2v^2$ for some integers $u,v$, or else $x+1 = 2u^2$ and $x-1=v^2$ for some integers $u,v$. Rather than repeatedly asking questions, maybe spending some - even minimal - amount of time in consideration, would help you.