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Topic: Integer Constant And Divisibility (Read 385 times) |
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K Sengupta
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Integer Constant And Divisibility
« on: Jun 5th, 2008, 7:34am » |
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C is a positive integer such that both (C*m + 1) and C*(m+1) + 1 are perfect squares, where m is a positive integer constant.
Prove that C is always divisible by 8(2m+1).
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Eigenray
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Re: Integer Constant And Divisibility
« Reply #1 on: Jun 5th, 2008, 9:53am » |
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Some thoughts:
write mC+1 = r2, (m+1)C+1 = s2, h=s-r. Then
(r - mh)2 = m(m+1)h2 + 1.
If (m, h, r, C) is a solution, with C = h(2r+h)) and r2 = mC+1, then (m, h', r', C') is a solution, where h' = h+2r, r' = r+2mh', C' = h'(h'+2r'). So there is an infinite family of parameterizations:
h0(m) = 0; r0(m) = 1; C0(m) = 0.
h1(m) = 2; r1(m) = 4m+1; C1(m) = 16m + 8.
h2(m) = 4(2m+1); r2(m) = 16m2+12m+1; C2(m) = (8m+4)(32m2 + 32m + 6)
h3(m) = 2(4m+1)(4m+3); r3(m) = 64m3 + 80m2 + 24m + 1;
h4(m) = 8(2m+1)(8m2+8m+1); r4(m) = (4m+1)(64m3 + 96m2+36m+1)
...
hn = hn-1 + 2rn-1; rn = rn-1 + 2mhn; Cn = hnhn+1.
Since
h'' = h' + 2r'
= h+2r + 2(r+2m(h+2r))
= (4m+1)h + 4(2m+1)r,
and 4(2m+1) | h0, it follows 4(2m+1) | hn for all even n; and since 2|h1, it follows 2|hn for all odd n. Therefore Cn = hnhn+1 is divisible by 8(2m+1) for all n,m.
So it suffices to show that this family contains all solutions. A program suggests that this is true.
We may solve the above recurrence to get
hn = [ (1+2m+2x)n - (1+2m-2x)n ]/(2x),
rn = [ (x-m)(1+2m-2x)n + (x+m)(1+2m+2x)n ]/(2x),
where x =  (m(m+1)).
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| « Last Edit: Jun 5th, 2008, 10:53am by Eigenray » |
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Eigenray
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Re: Integer Constant And Divisibility
« Reply #2 on: Jun 5th, 2008, 10:55am » |
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Oh. This is actually just Pell's equation. We need only check that (2m+1, 2) is the fundamental solution to x2 - m(m+1)y2 = 1.
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