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Topic: integer solution (Read 440 times) |
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fatball
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Can anyone help me think outside the box please?
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integer solution
« on: Jan 22nd, 2006, 7:40pm » |
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Find all solutions to c2 + 1 = (a2 - 1)(b2 - 1), in integers a, b, and c.
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inexorable
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Re: integer solution
« Reply #1 on: Jan 22nd, 2006, 7:59pm » |
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I see infinite solutions with b!=-1,1 and a=b^2
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Eigenray
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Re: integer solution
« Reply #2 on: Jan 22nd, 2006, 11:32pm » |
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on Jan 22nd, 2006, 7:59pm, inexorable wrote:| I see infinite solutions with b!=-1,1 and a=b^2 |
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I see no such solutions. For then
c2 = b6 - b4 - b2 = b2(b4 - b2 - 1),
and the second factor is never a square, being 3 mod 4.
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Eigenray
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Re: integer solution
« Reply #3 on: Jan 23rd, 2006, 4:09am » |
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| hidden: | We can write the equation as
a2(b2-1) = b2+c2.
Now, if b is odd, then the LHS is divisible by 4, while the RHS is either 1 or 2 mod 4. So b is even, and b2-1 = 3 mod 4.
If a were odd, then the LHS would be 3 mod 4, while the RHS can't be. Then a is even, making the LHS even, forcing c even also. Setting a',b',c' = a/2, b/2, c/2, we get
a'2(4b'2-1) = b'2+c'2.
Again, a' can't be odd for the same reason as before, and a' even makes the LHS divisible by 4, forcing b',c' even also. Now a'',b'',c'' = a/4, b/4, c/4 satisfy
a''2(42b''2-1) = b''2+c''2,
which again forces a'',b'',c'' all to be even just as above. Repeating this argument shows a,b,c to be divisible by arbitrarily large powers of 2, which means a=b=c=0. |
[edit]Alternate solution: (a2-1)(b2-1)=c2+1
| hidden: | | If b were odd, the LHS would be divisible by 4, while the RHS can't be. So b is even, and b2-1 =3 mod 4. If b != 0, then the LHS must be divisible by some prime q=3 mod 4. But c2 = -1 mod q is impossible, as that would imply c has order 4 in the group (Z/q)*, of order q-1, but 4 doesn't divide q-1. This contradiction means b=0, from which it follows a=c=0. |
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| « Last Edit: Jan 23rd, 2006, 4:50am by Eigenray » |
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