Determine all possible values of a prime number P such that P³-4P+9 is a perfect square.

Note: Computer program aided solutions are welcome, but a semi-analytic solution is preferred.

Write the expression as P(P^2 - 4) + 9 = P(P+2)(P-2) + 9

P, P+2, and P-2 are all different mod 3 and so one of them must = 0 mod 3 and the product is a multiple of 3. That means the sum is also a multiple of three (since 9 is) and if it’s a perfect square must be a multiple of 9. Since 9 already is, that means one of P, P-2, and P+2 must be a multiple of 9 (exactly one of them can contribute a factor of 3.) But P can’t be a multiple of 9 and also prime, so exactly one of P-2 and P+2 is a multiple of 9.

Separate out the case of P = 2, which is a solution since (2)(0)(4) + 9 = 9 ,which is a perfect square. 

Otherwise, P, P-2 and P+2 are all odd and therefore the one that’s a multiple of 9 is an odd multiple of 9. Write it as (2k+1)*9

So P is (2k+1)*9 +- 2 = 18k + 9 +- 2. Write the expression in terms of k:

 18^3k^3 + 3*18^2k^2(9 +- 2) + 3*(18*k)(9 +- 2)^2 + (9+-2)^3 - 4*18k - 4(9 +-2) + 9

= 

(18^3)k^3 + (3*18^2*(9 +-2)k^2 + (3*18*(9+-2)^2 - 4*18)k + ((9+-2)^3 - 4(9+-2) + 9)

All of these terms are even regardless of k so if the sum is a perfect square they must all be multiples of 4 or else there must be an even number of terms only divisible by 2. The first term is divisible by 8 and the second by 4, and the third either 4 or 32 depending on whether we use the + or the -; since these three terms are all divisible by 4, the linear term must be divisible by 4 as well. Unfortunately, it’s not since its first term has only one factor of two and its second is divisible by 4. 

Since there are no cases where this expression’s coefficients allow for a perfect square, the only other possibility is that the coefficients don’t matter because k itself is zero. In THAT case, the first three terms vanish and we’re left with ((9+-2)^3 - 4(9+-2) + 9) being a perfect square. As it happens, in both the + and - case, this is the case, as can be verified manually.

So the cases of P=2,7, and 11 all yield solutions, and there can be no others.

Posted by Paul on 2022-12-12 15:31:33