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Why is this allways true?

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Counter: 4^5 + 5^4 = 1649. 1648/64 = 25.75.
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The result is not always an integer, but at least always a multiple of a quarter.  
Assuming x>=2, then x\^(x+1) is clearly divisible by x², and if you use the binomial theorem to expand (x+1)\^x then you'll find that every summand except the 1 is dividible by x². The 1 is subtracted, so the result is divisible by x². You are trying to divide by 4x², and that extra division by 4 doesn't always work, so you can end up not only with integers but also with .25 of .5 or .75.
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For your edit, x will always divide x^(x+1), and (x+1)^x - 1 is a difference of powers which equals (x+1)^x - 1^x = (x) * ∑(x+1)^(i) from i = 0 to i = x-1 meaning x divides ((x+1)^x - 1). x^(x+1) + (x+1)^x - 1 will always be even thus it is also divisible by 2x.
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We can prove something nearly as good: for positive integers x, x² divides f(x):=x^{x+1}+(x+1)^x -1. Since x+1≥2, it's immediately obvious x² divides x^{x+1}. Moreover, when we expand (x+1)^x with the binomial expansion theorem, we'll get a whole bunch of terms divisible by x², plus the remaining terms xC1 * x + xC0 * 1, which simplifies to x²+1. So when we finally take off 1, we'll get something divisible by x². Furthermore, by considering the cases x=0,1,2,3 mod 4 separately, for x>1 we deduce that f(x) is divisible by 4. Therefore, _if x is odd and >1_ we can conclude f(x) is divisible by 4x²=(2x)², as required. But if x is even, this needn't hold, a another commenter has already shown - we can only deduce divisibility by x².

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