Delicate prime pair

We must consider 3 cases in this problem.

Case 1: Assume that $p$ and $q$ be primes greater than 3.

If $p$ and $q$ are primes greater than 3, then the two primes must be $= 1(mod 3)$ or $= 2(mod 3)$ . Now, we will have 4 sub cases.

Sub case 1: Assume that $p = 1(mod 3) ; q = 1(mod 3)$

By Substitution and Properties of Modulo:

$p^3 - q^5 = (p + q)^2$

$1^3 - 1^5 = (1 + 1)^2$

$1 - 1 = 2^2$

$1 - 1 = 4$

$1 - 1 = 1$ (Since $4 = 1(mod 3)$ )

$0 = 1$

which is false so our assumption is incorrect

Sub case 2: Assume that $p = 1(mod 3) ; q = 2(mod 3)$

By Substitution and Properties of Modulo:

$p^3 - q^5 = (p + q)^2$

$1^3 - 2^5 = (1 + 2)^2$

$1 - 32 = 3^2$

$1 - 32 = 9$

$1 - 2 = 0$ (Since $32 = 2(mod 3) ; 9 = 0(mod 3)$ )

$-1 = 0$

which is also false so our assumption is incorrect

Sub case 3: Assume that $p = 2(mod 3) ; q = 1(mod 3)$

By Substitution and Properties of Modulo:

$p^3 - q^5 = (p + q)^2$ $2^3 - 1^5 = (2 + 1)^2$ $8 - 1 = 3^2$ $8 - 1 = 9$ $2 - 1 = 0$ (Since $8 = 2(mod 3) ; 9 = 0(mod 3)$ ) $1 = 0$

which is also false so our assumption is incorrect

Sub case 4: Assume that $p = 2(mod 3) ; q = 2(mod 3)$

By Substitution and Properties of Modulo:

$p^3 - q^5 = (p + q)^2$

$2^3 - 2^5 = (2 + 2)^2$

$8 - 32 = 4^2$

$8 - 32 = 16$

$2 - 2 = 1$ (Since $16 = 1(mod 3)$ )

$0 = 1$

which is also false so our assumption is incorrect

Therefore, our assumption in case 1 is incorrect.

Case 2: let $p$ be a prime less than or equal to 3

$2$ and $3$ are the only primes less than or equal to 3

Sub case 1: p = 2

By substitution using $p = 2$ :

$p^3 - q^5 = (p + q)^2$

$2^3 - q^5 = (2 + q)^2$

$8 - q^5 = 4 + 4q + q^2$

$0 = q^5 + q^2 + 4q - 4$

All of the roots of $q^5 + q^2 + 4q - 4 = 0$ are not rational by the use of Rational Root Test

Therefore, our assumption is incorrect

Sub case 2: p = 3

By substitution using $p = 3$ :

$p^3 - q^5 = (p + q)^2$

$3^3 - q^5 = (3 + q)^2$

$27 - q^5 = 9 + 6q + q^2$

$0 = q^5 + q^2 + 6q - 18$

All of the roots of $q^5 + q^2 + 6q - 18 = 0$ are not rational by the use of Rational Root Test

Therefore, our assumption is incorrect

Case 3: let $q$ be a prime less than or equal to 3

Sub case 1: q = 2

By substitution using $q = 2$ :

$p^3 - q^5 = (p + q)^2$

$p^3 - 2^5 = (p + 2)^2$

$p^3 - 32 = p^2 + 4p + 4$

$p^3 - p^2 - 4p - 36 = 0$

All of the roots of $p^3 - p^2 - 4p - 36 = 0$ are not rational by the use of Rational Root Test

Sub case 2: q = 3

By substitution using $q = 3$ :

$p^3 - q^5 = (p + q)^2$

$p^3 - 3^5 = (p + 3)^2$

$p^3 - 243 = p^2 + 6p + 9$

$p^3 - p^2 - 6p - 252 = 0$

One of the root of $p^3 - p^2 - 6p - 252 = 0$ is rational which is $7$ and $7$ is also a prime therefore $(p,q) = (7,3)$ satisfies all the given conditions.

Therefore, the sum of all primes $p$ which is also the only possible value of $p$ is $\boxed{7}$ .

The question is not as difficult as it seems to be. The equation clearly shows p and q cannot be very large values , so you need to try small values of primes. The smallest pair which satisfies the equation is , p= 7 and q=3.Above these values , the difference between the LHS and RHS of equation is highly unequal, so you need not check further. so, we have only one value of p satisfying the equation, i.e. 7, and that is the answer.But the question is ,no doubt, interesting.

Let's make a table in mod 3.

• p q p^3-q^5 (p+q)^2 possibility
• 0 0 0 0 T
• 0 1 -1 1 F
• 0 -1 1 1 T
• 1 0 1 1 T
• 1 1 0 1 F
• 1 -1 -1 0 F
• -1 0 -1 1 F
• -1 1 1 0 F
• -1 -1 1 1 T
  Trial and error shows that only possible ans is 7,3.