Prime Pairs satisfy Quintic = Cubic (Posted on 2024-02-17)

        p(p4+p2+10q) = q(q2+3)

	Submitted by K Sengupta
Rating: 5.0000 (1 votes)
Solution:	(Hide)
If p=q, then we have lhs > rhs. Accordingly, it follows that p ≠ q. If p>3, then p| q^2+3, and hence -3 is a quadratic residue in (mod p) and consequently: p ≡ 1(mod 3). Then, looking at modulo 3, we get: p(p^4+p^2+10q) ≡ q+2(mod 3) while, q(q^2+3) ≡ q(mod 3) By Fermat's Little Theorem, or by direct check, this leads to a contradiction. Checking the values p=2,3, we find that: the only solution is (p,q) = (2,5).

	Subject	Author	Date
	No Subject	K Sengupta	2024-02-22 23:33:03
	Woah	Daniel Adams	2024-02-22 15:18:40
	re(2): Analytic solution with computer verification	Larry	2024-02-18 09:23:00
	p=2	xdog	2024-02-17 21:50:59
	re: Analytic solution with computer verification	Brian Smith	2024-02-17 19:04:00
	Analytic solution with computer verification	Larry	2024-02-17 18:34:10
	probable solution	Charlie	2024-02-17 15:52:16