Let $p=2^a1>7$ be a Mersenne prime and so $a$ is an odd prime. Can we say that $(p^2+1)/2$ is not equal to the square of a prime number? Many thanks for your help BHZ

$\begingroup$ Oh I am sorry I did not think to this relation. I need it in character theory of finite group and this shows that this is impossible. Thank you very much $\endgroup$– BHZCommented May 23, 2013 at 9:16

$\begingroup$ Doesn't looking modulo 3 show that $a$ has to be odd? The equation is $2^a \equiv p_0^2 (\text{mod } 3)$ and assuming $p_0 \neq 3$ we have $2^a \equiv 1 (\text{mod } 3)$. Any odd $a$ would satisfy this. $\endgroup$– rghthndsdCommented May 23, 2013 at 12:13

$\begingroup$ You are absolutely right. A stupid sign error. I delete my comment, so ashamed I am of it. $\endgroup$– Chris WuthrichCommented May 23, 2013 at 13:04
2 Answers
Suppose, $p^22p_1^2=1.$ Substituting $p=2^a1,$ we arrive at $$2^a(2^{a1}1)=(p_11)(p_1+1).$$ Observe, $(p_11,p_1+1)=2,$ so we must have the following options: $p_11=2^{a1}k$ and $p_1+1=2l$ and $kl=2^{a1}1.$ This is impossible unless $k,l$ and thus $a$ are small. Indeed, if $k\ge 2,$ then $p_1\ge 2^a+1$ and $l\ge 2^{a1}$ so $kl\ge2^a.$ If $k=1,$ then $p_1=2^{a1}+1$ and $l=2^{a2}+1$ and $kl=2^{a2}+1=2^{a1}1.$ This implies $a=3.$ Otherwise, $p_11=2k$ and $p_1+1=2^{a1}l$ and $kl=2^{a1}1.$ Again it is possible only if $k,l$ and $a$ are small. These case can be checked by hands.

$\begingroup$ I find it more clea to say "In the first case,$l = 2^{a2}k+1$, so $2^{a2}k^2 + k = 2^{a1}1$, thus $k^2 \lt 2$ and $a \lt 4$.", and come up with an analogous expression and similar bound on $a$ in the second case. Nice argument. Gerhard "Likes Working With Small Numbers" Paseman, 2013.05.23 $\endgroup$ Commented May 23, 2013 at 16:39

$\begingroup$ Clear, not clea. Also, the second case leads to $2^{a2}(l^2  2) = l  1$, from which one notes that $a \gt 3$ gives no positive integer solutions for $l$. Gerhard "Also Likes Working With Inequalities" Paseman, 2013.05.23 $\endgroup$ Commented May 23, 2013 at 16:45

$\begingroup$ I am really thankful from all of you for your helps. $\endgroup$– BHZCommented May 24, 2013 at 16:19
A result of Szalay [Indag. Math. 2002] gives you what you want. He proves that the equation $$ 2^x2^y+1 = z^2 $$ has only the solutions $$ (x,y,z)=(2t, t+1, 2^t1), (t1,t1,1), (5,3,5), (7,3,11), (15,3,181) $$ in positive integers $x, y$ and $z$ (here, $t \geq 2$ is an integer). The proof uses lower bounds for $$ \left \sqrt{2}  \frac{p}{2^k} \right $$ and elementary arguments.