Let p(p2-p-1)=q(2q+3) with p,q, prime
p does not divide q, so p must divide (2q+3), say a times.
(2q+3)=a*p,p=(2q+3)/a
Substituting:
((2q+3)/a)(((2q+3)/a)^2-((2q+3)/a)-1)=q(2q+3)
Less of a mess than it looks, since (2q+3)^2 = a(a^2q+a+2q+3),
Evaluate RHS for small values of a:
0,
(3q+4),
2(6q+5)
3(11q+6)
4(18q+7)
5(27q+8)
...
(2q+3)^2=5(27q+8), with q=31 is the sole positive integer value for q (so a is also prime).
But then p=(2q+3)/a= (62+3)/5=13 must be the sole positive integer value for p
So the answer is {p,q}={13,31}
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Posted by broll
on 2023-11-17 22:47:59 |
Possible Solution
