And we already know a in terms of d. There's probably a more general relationship I missed by making the assumptions c = 1 and a = d+2, but this is not the math I'm being paid to do. :)
you know, i listen to the odd-numbered symphonies now and then but i hardly ever listen to the even-numbered symphonies. my favorites are 3 and 7, but maybe i need to go back to the evens for a bit.
(that's about as mathematical as my thinking ever gets, btw.)
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27^2 * 6 + 1 = 25^2 * (6 + 1)
31^2 * 7 + 1 = 29^2 * (7 + 1)
35^2 * 8 + 1 = 33^2 * (8 + 1)
39^2 * 9 + 1 = 37^2 * (9 + 1)
and so on...
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(a^2)b+c = (d^2)(b+c)
rearranged to:
(a^2-d^2)(b+c) = c(a^2-1)
Decided c = 1 would make life easier:
(a^2-d^2)(b+1) = (a^2-1)
Wanted to make the leftmost factor simpler, so I set a = d+2 (based on your example):
(4d+4)(b+1) = d^2+4d+3
4(d+1)(b+1) = (d+1)(d+3)
4(b+1) = d+3
4b+1 = d
And we already know a in terms of d. There's probably a more general relationship I missed by making the assumptions c = 1 and a = d+2, but this is not the math I'm being paid to do. :)
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(that's about as mathematical as my thinking ever gets, btw.)