Prime Equation
Let p and q be primes such that p + q = ( p − q ) 3 . What is the sum of all possible values of p and q ?
The answer is 8.
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1 solution
is there no theoretical way of solving it?
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Consider the equation ( m o d 3 ) . If neither p or q is 3 then both must be either 1 or 2 ( m o d 3 ) . Evidently if both have the same remainder ( m o d 3 ) then the right hand side is divisible by 3 but the left hand side is not. If both have different remainders ( one of them is 1 and the other is 2) then the left hand side is divisible by 3 but the right hand side is not. Therefore one of the primes must be 3. The left hand side is always positive, so p > q . So, if p = 3 then q = 2 but this is not a valid solution. If q = 3 then we seek all solutions to p + 3 = ( p − 3 ) 3 . So 3 ≡ − 2 7 ( m o d p ) , therefore 3 0 ≡ 0 ( m o d p ) . Therefore p can only be 2, 3 or 5. But p > q , so p = 5 is the only feasible solution, and this does indeed yield a solution. So ( 5 , 3 ) is the only solution
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Alternatively, if their difference is odd, then q must be 2. This leaves fairly clearly no solutions, as the RHS is larger than the left for p>3. Otherwise, p-q = 2x.
8x^3 = p+q=q+2x+q=2x+2q, so 4x^3 = x+q, so q=x(4x^2 -1), which is prime, so x=1. This implies that p-q = 2 and so p+q = 8, leaving (5,3) as the only solution.
Lets consider the difference for 2^3=8, 4^3=64, 6^3=216. So more the difference more should be the value and for those large values even the consecutive prime numbers generates a bit large value of difference which is when cubed would fetch even larger values. So we can try it with 8, which is 5 & 3, also it satisfies the equation.