prove it by number theory?

Question

(a1)^(n)+(a2)^(n)+.....+ (aM)^(n) can be expressed as difference of at least one set of two perfect squares.a1 to aM are natural and rational numbers and the resulting number from the summation must not have a single 2 as its multiple as single 2 canot be expressed as difference of two perfect squares.

Answer 1

You don't need to limit yourself to numbers of the form (a1)^(n) + (a2)^(n) + ... + (aM)^(n)

Let S = A^2 - B^2 = (A + B) (A - B) = N M with N = A + B, M = A - B.

Solving for A and B yields

A = (N + M) / 2
B = (N - M) / 2

For A and B to be integral you just need N + M and N - M to both be even. This happens if and only if N and M have the same even/odd parity: so either N and M are both odd (and their product S = N M is also odd) or both N and M are even (and their product S = N M is divisible by 4). So if an integer S is a difference of two squares of integers then either S is odd or S is a multiple of 4.

Conversely, if S is odd then N = S, M = 1 are both odd and if S is divisible by 4 then N = S / 2, M = 2 are both even, so for S odd or divisible by 4 we have a solution S = A^2 - B^2 with A = (N + M)/2 and B = (N - M)/2.

Thus integers S are the difference of two squares of integers if and only if S is odd or a multiple of four. Integers S are not the difference of two squares if and only if S is "oddly even" i.e., divisible by two but not by four -- { +/- 2, +/- 6, +/- 10, ... }. In the latter case however there are rational solutions, for example 2 = (3/2)^2 - (1/2)^2. But then of course every rational is a difference of two squares of rationals: q = ((q + 1)/2)^2 - ((q - 1)/2)^2.

Answer 2

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Answer 3

Assume all aj are integers, since we can multiply by a large square to clear fractions.
We have a1^n + a2^n + ... = c
If c is odd, we have
x^2 - y^2 = c
let x + y = c
x - y = 1
x = (c+1)/2
y = (c-1)/2
If c is even, then we must factor c into d*f
(x-y)(x+y) = d * f
For this to work, we must have d and f being both even or both odd. since x-y and x+y have the same parity. This will work exactly when c is divisble by 4 and not just by 2.

Answer 4

x=2...x always equals 2
i get the points,right?