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Old 2008-05-03, 23:20   #3
wblipp
 
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"William"
May 2003
New Haven

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Zeta-Flux has already observed that a cannot be even (he said it was easy to see, but it took me a while to see that it's because x and y and z must all be odd by the gcd requirement, and xy + xz + yz would then be odd)

let a = 2b+1

x = -(2b+1)^2 * c^2 * (18b+10) - 3*(2b+1)*c*(b+1) + b
y = (2b+1)^2 * c^2 * (54b+30) - (2b+1)*c*(9*b+1) + 1
z = (2b+1)^2 * c^2 * (27b+15) + (2b+1)*c*(9*b+7) + 1

Gives an infinite family of solutions, and the gcd requirement is met because

x = b mod (2b+1)
y = 1 mod (2b+1)
z = 1 mod (2b+1)

Thus there is an infinite family of solutions for all odd a.

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