Fermat showed that no two fourth powers can ever sum to a perfect square.
Go one better by proving that there is no solution in the positive integers for the equation a^4+b^4+1 = c^2.
Or give a counter-example.
(Acknowledgement: suggested by a post of Daniel's on one of my earlier problems)
If one or two of a,b is odd, then LHS = 2 or 3 mod 4. But all squares = 0 or 1 mod 4, so both a and b are even.
The terminal digit of even 4th powers is 0 or 6, so the terminal digit of LHS is 1, 7, or 3. But no square ends in 7 or 3, so both a and b are multiples of 10, c is odd, and c^2 = 1 mod 10000.
Setting a = 10x, b = 10y, c = 2z + 1 gives 2500(x^4 + y^4) = z(z + 1) and z is divisible by 625 or 624.
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Posted by xdog
on 2010-12-27 10:38:26 |
A start
