Primes and Squares

Suppose that $13p + 4 = x^{2}$ . Then $13p = (x+2)(x-2)$ . One of the two brackets on the right hand side must be divisible by 13. We have 2 cases: Firstly, $x+2 = 13m$ for some positive integer $m$ : $13p=13m(13m-4)$ $p = m(13m-4)$ $13m - 4 > m$ so this can only be possible if $m = 1$ , but that would mean that $p = 9$ which is not prime

Now we must consider the other case, $x-2 = 13n$ for some positive integer $n$ : $13p=13n(13n+4)$ $p = n(13n+4)$ Again, this can only be the case if $n = 1$ , meaning that $p = 17$ which is indeed a solution. Therefore the only solution and thus sum of all solutions is $\fbox{17}$

Say $13p + 4 = k^{2}$ . Then $13p = (k + 2)(k - 2)$ . As $p$ is prime, we have four possible cases:

$A)$ . $k + 2 = 13 \rightarrow p = 9$ .

$B)$ . $k - 2 = 13 \rightarrow p = 17$ .

$C)$ . $k - 2 = 1 \rightarrow p = \frac {5}{13}$ .

$D)$ . $k + 2 = 1 \rightarrow p = \frac {-3}{13}$ .

Clearly we only have $\boxed{17}$ as the answer since it's the only prime number.

Let 13p + 4 = k2 Therefore, 13p = (k +2)(k – 2). Then, we are left with only two possible cases :

1). k +2 = 13, then p = 9

2). k – 2 = 13 then p= 17

Hence, in this only 17 is a prime number. Therefore sum = 17

13p+4=K^2 =>13p=(k+2)(k-2) Let 13=(k+2)(a/b) Such that a & b are co-primes So p=(k-2)(b/a) =>b divides k+2 & a divides k-2 If p=k-2 & 13=k+2 or viceversa =>p=17 or p=9 But p must be a prime Hence p=17;

by question..... 13p+4 = \n^2\ or, 13p=\n^2-4 or, p={(n-2)(n+2)}/13 since, n-2 and n+2 are integers, either n-2=13k or n+2= 13k where k is an integer.... however, since p is a prime number... either n-2=13 or n+2 =13. when we put n=11, we dont get a prime number whereas if we put 17, we do.... so 17 is the only possible solution.