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Close to Fermat (Posted on 2007-04-06) Difficulty: 3 of 5
Show that if an+bn= 2m, and a, b, m, and n are positive integers (n>1), then a=b.

See The Solution Submitted by Federico Kereki    
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Solution Solution To The Problem Comment 4 of 4 |

We write a = (2^r)*p and b = (2^s)*q, for some non-negative integers
r and s, and odd integers p and q.
Accordingly, we obtain:(2^r*(p))^n + (2^s*(q))^n = 2^m.

We can assume without any loss of generality that  r< = s, giving:
 p^n + (2^(s-r)*(q))^n = 2^(m-nr).

Since both terms on the left-hand side of the equation are positive
integers, the right-hand side must be greater than one, and hence even.

Then,  p^n odd would give, (2^(s-r)*q)^n odd , so that r = s.

Hence we may cancel the factor 2^r from a and b, obtaining :
p^n + q^n = 2^t, for some positive integer t.

Case A: n is odd

We assume that  at least one of p and q are different from  1.
Since n is odd, p + q is a factor of p^n + q^n.

If C = p^n + q^n = 2^t, then
A = p + q = 2^u, for some positive integer u, and
B =
p^(n-1) - p^(n-2)q + p^(n-3)q^2 - p(n-4)q^3 + ... + q^(n-1)
= 2^(t-u),
and clearly, since C > A, we have B > 1, t > u, and so B is even.

But then B is the sum of an odd number of odd terms, and so is odd,
a contradiction.

Case B: n is even

Writing n = 2w, we have (p^w)^2 + (q^w)^2 = 2^tt, and it will be
sufficient to show that we must have p^w = q^w = 1, implying p = q = 1.

Since p^w and q^w are odd, their squares are congruent to 1(Mod 4).

Hence we have 1 + 1 = 2t (mod 4), implying t = 1.

Hence p^w = q^w = 1, and p = q = 1.

In both cases we conclude that p = q = 1, which is indeed a solution
of p^n + q^n = 2^t, with t = 1.

Hence, if a^n + b^n = 2^m, a = b = 2^r is a solution. (With nr + 1 = m.)

Therefore, a^n + b^n = 2^m  gives a = b, as was asked for.


  Posted by K Sengupta on 2007-04-09 13:37:11
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