$p$ has to be $3$ and no other prime is valid in the above equation....

sarthak7 wrote: $p$ has to be $3$ and no other prime is valid in the above equation.... can you explain

t $x^2=p^2+7^n$ now we have $x-p=7^a$ and $x+p=7^b$ so $2p=7^b-7^a$ so we have case 1: $7|p$ and then$ p=7$ so $a=1$ and then $7^b=21$ which can not be true case 2: $a=0$ so now$2p=7^b-1$ so $3|p$ so $p=3$ and $n=1$

nikolapavlovic wrote: t $x^2=p^2+7^n$ now we have $x-p=7^a$ and $x+p=7^b$ so $2p=7^b-7^a$ so we have case 1: $7|p$ and then$ p=7$ so $a=1$ and then $7^b=21$ which can not be true case 2: $a=0$ so now$2p=7^b-1$ so $3|p$ so $p=3$ and $n=1$ how can you assume $x-p=7^a$ and $x+p=7^b$, it may be either of one is 1.

$7\equiv 1\mod 3$ But $p^2$ is of the form $3k+1$ except for $3$ So, to be perfect square $p$ has to be $3$

No, solution exists as $7^n$ is congruent to $1$ modulo $6$ p is a prime then it can be $\pm1$ modulo 6 but $p^2$ will always be 1 modulo 6

I have not told solution doesnt exist....

Chirantan wrote: nikolapavlovic wrote: t $x^2=p^2+7^n$ now we have $x-p=7^a$ and $x+p=7^b$ so $2p=7^b-7^a$ so we have case 1: $7|p$ and then$ p=7$ so $a=1$ and then $7^b=21$ which can not be true case 2: $a=0$ so now$2p=7^b-1$ so $3|p$ so $p=3$ and $n=1$ how can you assume $x-p=7^a$ and $x+p=7^b$, it may be either of one is 1. $(x-p)(x+p)=7^n$

nikolapavlovic wrote: Chirantan wrote: nikolapavlovic wrote: t $x^2=p^2+7^n$ now we have $x-p=7^a$ and $x+p=7^b$ so $2p=7^b-7^a$ so we have case 1: $7|p$ and then$ p=7$ so $a=1$ and then $7^b=21$ which can not be true case 2: $a=0$ so now$2p=7^b-1$ so $3|p$ so $p=3$ and $n=1$ how can you assume $x-p=7^a$ and $x+p=7^b$, it may be either of one is 1. $(x-p)(x+p)=7^n$ it may be $(x-p)=7^n$ and $(x-p) = 1$, you never know

Chirantan wrote: No, solution exists as $7^n$ is congruent to $1$ modulo $6$ p is a prime then it can be $\pm1$ modulo 6 but $p^2$ will always be 1 modulo 6 $p=3$,$n=1$

[/quote] $(x-p)(x+p)=7^n$ it may be $(x-p)=7^n$ and $(x-p) = 1$, you never know[/quote] well that s just $b=0$ look at the solution

nikolapavlovic wrote: Chirantan wrote: No, solution exists as $7^n$ is congruent to $1$ modulo $6$ p is a prime then it can be $\pm1$ modulo 6 but $p^2$ will always be 1 modulo 6 $p=3$,$n=1$ I forgot to mention it that the only solution is that as 3 is the only prime not of the form $6k\pm1$

Yes the only solution is $p=3$ and $n=1$. We obtain $p=3$ by considering the equation modulo 3 and the fact that $x^2=p^2+7^n$ is equal to 0 or 1 modulo 3. After we solve the equation $(x-3)(x+3)=7^n$ so both $x-3$ and $x+3$ are power of 7 with difference equal to 6. Thus $n=1$.

Claim: $\boxed{(n,p)=(1,3)}$ is the only solution. Proof: Let: $p^2+7^n=k^2$. Notice that for $p\ge 5$ and $p=2$ we have that $p^2+7^n\equiv 2 \pmod 3$ which is a contradiction, since $k^2\equiv 0,1 \pmod 3$, thus $p=3$. Thus $9+7^n=k^2\Longrightarrow 7^n-1^n=k^2-10$, furthermore by $LTE, \nu_3(7^n-1^n)=\nu_3(n)+\nu_3(6)=\nu_3(k^2-10)$. FTSOC, assume $\nu_3(n)\neq 0 \Longleftrightarrow n=3t$ thus the equation becomes: $9+343^t=k^2\Longrightarrow 9+343^t\equiv 9 \pmod {343} \Longleftrightarrow k^2\equiv 9 \pmod {343}\Longrightarrow k^2=343x+9 \Longleftrightarrow 343=\frac{k^2-9}{x}$. Furthermore $x=343^{k-1}\Longrightarrow x=(k^2-9)^{1-\frac{1}{k}} \Longrightarrow$ $k=\log_{343} (343^{1-\frac{1}{k}}x^{1-\frac{1}{k}})=\log_{343} (x^{1-\frac{1}{k}})-\frac{1}{k}+2$, however this is a contradiction since $RHS\notin \mathbb{Z}$. Thus $\nu_3(n)=0$ $\blacksquare$. This implies that $\nu_3(6)=\nu_3(k^2-10)\Longrightarrow k^2=16 \Leftrightarrow k=4, n=1$. QED. And we are done!

tchebytchev wrote: Find all positive integer $n$ and prime number $p$ such that $p^2+7^n$ is a perfect square $$p^2+7^n=x^2...(\alpha)$$$$\Rightarrow (x-p)(x+p)=7^n$$$$\Rightarrow x-p=7^a, x+p=7^b, a,b\in \mathbb{R}^+_0$$If $a,b\ge 1:$ $$\Rightarrow x-p\equiv x+p\pmod{7}$$$$\Rightarrow p\equiv 0 \pmod{7}$$$$\Rightarrow p=7$$$$\Rightarrow 7^b=7^a+14$$$$\Rightarrow 7^{b-1}=7^{a-1}+2 (\Rightarrow \Leftarrow)$$ $$\Rightarrow a=0$$$$\Rightarrow x=p+1$$In $(\alpha):$ $$2p=7^n-1, n\in \mathbb{N}$$If $n\ge 2:$ $$7^n-1\equiv 0 \pmod{3}$$$$\Rightarrow p=3$$$$7-1=2\times 3$$By Zsigmondy´s theorem: $$\exists \text{ prime }p \neq 2,3/ p|7^k+1, \forall k\in \mathbb{N}_{\ge 2}$$$$\Rightarrow p\neq 3 (\Rightarrow \Leftarrow)$$ $$\Rightarrow n=1$$$$\Rightarrow p=3$$ $$\Rightarrow (n,p)=(1,3)\text{ is the only solution}_\blacksquare$$

I just realized that zsigmondy's theorem was absolutely unnecessary