# $(a+b+c\cdots)\neq(a^{2}+b^{2}+c^{2}\cdots)$ given all distinct values for the variables?

Please note that the solution must not require more equations to solve as do the variables increase.

Apparently, $(a+b+c\cdots)\neq(a^{2}+b^{2}+c^{2}\cdots)$ seems pretty obviously to be true given that all variables $a,b,c\cdots$ are distinct natural numbers, there is no disproof I can think of (also please note that the number of variables on the RHS and LHS are same and equal in quantity, i.e. $(a+b) \neq (a^{2} + b^{2})$ or $(a+b+c+d) \neq (a^{2}+b^{2}+c^{2}+d^{2}$).

If this is indeed true, it becomes obvious that there is only one unique solution for all variables that satisfies both equations (only and only two are given, as the example towards the end of this post). Wolfram|Alpha does the job to a certain degree, except it doesn't recognize that all variables must have different values, and also counts permutations of values among the variables as possible solution.

So, how do I find a solution given two equations, as in;

$a+b+c+d+e=1837, a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=1234567$

BTW the solution is $a=66, b=79, c=169, d=628, e=895$, also, if the conjecture is false, are there any counterexamples (again, please remember that all variables are distinct).

Thanks to whomsoever has offered to help.

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For any natural number, $a^2\geq a$ with equality only when $a=1$ or $a=0$. So your statement is true assuming you have more than $1$ term – Thomas Andrews May 2 '12 at 13:28
Might be possible if you took cubes instead of squares into account. Anyone agree? – user30453 May 2 '12 at 14:12
@Anonym, $a+b+c+d=26$, $a^2+b^2+c^2+d^2=234$, $a^3+b^3+c^3+d^3=2366$ has more than one solution. – Gerry Myerson May 2 '12 at 22:47

$a+b+c=12$, $a^2+b^2+c^2=62$ has the solution $a=1$, $b=5$, $c=6$, but it also has the solution $a=2$, $b=3$, $c=7$.

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Note that if $a+b+c+d=2n$, then you can show that:

$$(n-a)+(n-b)+(n-c)+(n-d)=2n$$ $$(n-a)^2+(n-b)^2+(n-c)^2+(n-d)^2 = a^2+b^2+c^2+d^2$$

So if you start with the right $(a,b,c,d)$ with distinct values, you can get a different one with distinct values. For example, with $n=13$, you could take $(a,b,c,d)=(2,5,7,12)$ and therefore $(n-a,n-b,n-c,n-d)=(11,8,6,1)$

There's actually a statistical reason for this. Knowing the number of variables, the sum of those variables, and the sum of the squares of the variables, is the same as knowing the number of variables, the average of those variables and the variance of those variables.

If you start with $(a,b,c,d,...)$ then $(-a,-b,-c,...)$ has the same variance. So does $(n+a,n+b,...)$. Therefore so does $(n-a,n-b,n-c,...)$. Now choose $n$ so that the resulting sum: $(n-a)+(n-b)+... = a+b+c+...$. You can only find an integer $n$ which works if twice the average of the set $a,b,c,...$ is an integer.

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