Primes Making Perfect Squares

One solution is $p=2$ . If there are any more solutions, $p$ must be an odd number since it is prime. Let $p = 2m+1$ . Consider the expression modulo 12, it will have value 5. However, any square number modulo 12 will be equal to $0,1,4,9$ . Hence there are no other solutions other than 2, so the sum of all elements of $P$ is 2.

a perfect square x=n^2,then x%3=0||1,x=5^p+12^p,so x%3 must be 1.(5^p)%3=(-1)^p %3=1%3,so p must be even.but p is also a prime.so p can be only 2

Solution 1: Let $5^p+12^p =n^2$ . Consider modulo 3. Since $n$ is an integer, $n \equiv 0,1$ or $2 \pmod{3}$ , so $n^2 \equiv 0\cdot 0 \equiv 0 \pmod{3}, n^2 \equiv 1\cdot 1 \equiv 1 \pmod{3}$ or $n^2 \equiv 2\cdot 2 \equiv 1 \pmod{3}$ . Thus, $n^2 \equiv 0,1 \pmod{3}$ . Hence $5^p + 12^p \equiv 5^p \equiv 2^p \equiv 0, 1 \pmod{3}$ . Since $2^p \not \equiv 0 \pmod{3}$ , we must have $2^p \equiv 1 \pmod{3}$ . Since $2^{2k+1} \equiv (2^2)^k\cdot 2 \equiv 1^k\cdot 2 \equiv 2 \pmod{3}$ , this implies that $p$ must be even. The only even prime is 2, thus $p=2$ is the only possible solution. Checking this case, we see that $5^2 + 12 ^2 = 13^2$ is a perfect square, thus $P=\{2\}$ and the sum of the elements in $P$ is $2$ .

Solution 2: Let $5^p+12^p=n^2$ . Consider modulo 5. Since $n$ is an integer, $n \equiv 0,1,2,3$ or $4 \pmod{5}$ , so $n^2 \equiv 0\cdot 0 \equiv 0 \pmod{5},$ $n^2 \equiv 1\cdot 1 \equiv 1 \pmod{5},$ $n^2 \equiv 2\cdot 2 \equiv 4 \pmod{5},$ $n^2 \equiv 3\cdot 3 \equiv 4 \pmod{5}$ or $n^2 \equiv 4\cdot 4 \equiv 1 \pmod{5}$ . Thus, $n^2 \equiv 0, 1, 4 \pmod{5}$ . Hence $5^p + 12^p \equiv 12^p \equiv 2^p \equiv 0, 1, 4 \pmod{5}$ . It is impossible to have $2^p \equiv 0 \pmod{5}$ thus $2^p \equiv 1, 4 \pmod{5}$ . Let's check the remainders of $2^p$ . Since $2^4 \equiv 16 \equiv 1 \pmod{5}$ , we have $2^{4k} \equiv 1 \pmod{5}$ , $2^{4k+1}\equiv 2 \pmod{5}$ , $2^{4k+2}\equiv 4 \pmod{5}$ and $2^{4k+3}\equiv 3 \pmod{5}$ . Hence, $p=4k$ or $4k+2$ . Since these numbers are even, the only possible prime is $p=2$ . Checking this case, we see that $5^2 + 12 ^2 = 13^2$ , thus $P=\{2\}$ and the sum of the elements in $P$ is $2$ .

prime number only can divided by itself or 1. let the perfect square=x^2 5^p+12^p=x^2 (5^p+12^p)^1/2=x since p must perfectly divide 2. so answer is 2

When p = 2, $5^2$ + $12^2$ = $13^2$ , hence 2 belongs to set P. Now, we claim that all p>2 does not satisfy the condition that $5^p$ + $12^p$ is a perfect square. We prove this by first observing that for any natural number k, $k^2 \equiv 0 \pmod{3}$ or $k^2 \equiv 1 \pmod{3}$ . Then, by writing p = $3 + m$ , $m \geq 0$ , $12^{3+m}$ = $({3 \times 4})^{3+m}$ $\equiv 0 \pmod{3}$ , and $5^{3+m}$ = $125 \times 25^m$ $\equiv 2 \pmod{3}$ . Therefore, $5^{3+m}$ + $12^{3+m}$ $\equiv 2 \pmod{3}$ , implying that for all (p \geq 3)\, the sum is not a perfect square. Since P = {2}, sum of elements = 2.