A number theory problem by Nikola Alfredi

I did this way :

As $\displaystyle 9n + 7 = k^2 \ \ k \in \mathbb{Z}^+$ , thus $\displaystyle k^2 \equiv 7 \pmod {9}$ . Or by checking residues we can say...

$\displaystyle k \equiv 4 \pmod {9} \implies k = 9m + 4 \ \ m \in \mathbb{Z}^+$ . Then $\displaystyle 9n + 7 = (9m + 4)^2 \implies n = 9m^2 + 8m + 1$ .

Now $\displaystyle n + 6 = 9m^2 + 8m + 7$ is odd and as well prime . So $m$ must be even , hence for least even integer $m = 2$ , $n + 6 = 59$ or $n = 53$ .

Since $n+6$ is prime, therefore $n$ must be odd. Let $n=2m+1$ . Then $9n+7=2(9m+8)$ . So $m$ must be even. Let $m=2p$ . Then $9p+4$ must be a perfect square. The smallest possible value of $p$ for this to hold is $5$ . But then $n+6=2m+7=4p+7=27$ is not a prime . The next possible value of $p$ is $13$ . For this value of $p$ , $n+6=2m+7=4p+7=59$ is prime . Hence the required answer is $n=2m+1=4p+1=\boxed {53}$ .

Focus on $9n + 7$ :

$n$ = 10, not a perfect square

$n$ = 11, not a perfect square

$n$ = 12, not a perfect square

$n$ = 13, not a perfect sqaure

etc. (you will find other perfect squares during this section)

$n$ = 53, a perfect square

Now, when n is a perfect square use $n + 6$ :

$n$ = 18, not a prime (24)

$n$ = 53, a prime (59)

Therefore, the number of $n$ that satisifes both $n + 6$ = prime and $9n + 7$ = perfect square is 53