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| Additively Quadratic 2 (Posted on 2023-11-11) |
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Determine all possible positive integer triplet(s) (F, G, H) such that each of F and G is a prime with F ≤ G, and:
F(F+2) + G(G+2) = H(H+5)
The first several without a pattern
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 | Comment 2 of 4 | 
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I was not able to find a pattern, but here are the solutions limiting primes to under 1,000.
(3,5,5) (3,19,18) (7,7,9) (7,67,66) (13,19,22) (13,199,198) (41,157,161) (61,547,549) (67,131,146) (89,733,737) (101,607,614) (103,509,518) (131,751,761) (151,691,706) (163,859,873) (173,373,410) (223,709,742) (229,823,853) (367,811,889) (373,839,917) (467,787,914) (547,631,834) (607,727,946) (607,991,1161) (751,797,1094) (787,811,1129)
--------- primes = [i for i in range(1000) if isprime(i)] size = len(primes) ans = []
for i in range(size): F = primes[i] for j in range(i,size): G = primes[j] LHS = F*(F+2) + G*(G+2) H = (-5 + (25 + 4*LHS)**.5)/2 if H%1 == 0: ans.append([F,G,int(H)])
for a in ans: print('({},{},{})'.format(a[0],a[1],a[2]))
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Posted by Larry
on 2023-11-11 10:03:10 |
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