Sum of two 23rd powers as a power of 2

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Sum of two 23rd powers as a power of 2

Postby zrosnbrick » Sat Apr 15, 2006 3:46 pm

If p and q are positive integers and if p[sup]23[/sup] + q[sup]23[/sup] is a power of 2 prove that p = q
[zeta]([alpha] + [beta]i) = 0 [imp] [alpha] = 1/2
[pi] + e
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Postby quintic » Sun Apr 16, 2006 2:21 am

I may be missing a simpler approach, but here is an outline of a proof...

Writing p = 2[sup]r[/sup]a and q = 2[sup]s[/sup]b for some non-negative integers r and s, and odd integers a and b, we must have r = s, for otherwise, if (wlog) r < s, a[sup]23[/sup] + (2[sup]s-r[/sup]b)[sup]23[/sup] would be odd and greater than 1, a contradiction.

Hence we have a[sup]23[/sup] + b[sup]23[/sup] = 2[sup]n[/sup], where n is a positive integer.

Assume not both a and b are equal to 1.
If a[sup]23[/sup] + b[sup]23[/sup] = 2[sup]n[/sup], then n > 23, and
A = a + b = 2[sup]m[/sup], for some positive integer m > 1, and
B = a[sup]22[/sup] - a[sup]21[/sup]b + a[sup]20[/sup]b[sup]2[/sup] - a[sup]19[/sup]b[sup]3[/sup] + ... + b[sup]22[/sup] = 2[sup]n-m[/sup], and clearly n - m > 1.

As a + b is divisible by 4, a = +/- 1, b = -/+ 1 (mod 4). But then B = -1 (mod 4), a contradiction.

Hence a = b = 1, implying p = q = 2[sup]r[/sup].
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Postby zrosnbrick » Sun Apr 16, 2006 2:31 am

Could you explain why that's a contradiction?
[zeta]([alpha] + [beta]i) = 0 [imp] [alpha] = 1/2
[pi] + e
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Postby quintic » Sun Apr 16, 2006 2:38 am

Do you mean why B = -1 (mod 4) is a contradiction? Because B = 2[sup]n-m[/sup], where n - m > 1, is a multiple of 4.

<added>
Looking at this again, I see that all we need is n > m (follows from a[sup]23[/sup] + b[sup]23[/sup] > a + b), so that B is even. But then B is the sum of an odd number of odd terms and so is odd. Contradiction.
</added>
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