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100 numbers are written on a board. After each number was increased by 1, the sum of their squares did not change. Will the sum of squares change and by how much if the numbers are increased by 1 again?
I don't know where to start since I can't find an example and that logically the sum will always change because the squares will always be non-negative.

Gerard L.
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    The squares may be non-negative, but are the numbers themselves? For example $(-4)^2+(3)^2 = (-4+1)^2+(3+1)^2$ – Henry Jan 27 '18 at 01:17
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    " logically the sum will always change " The set ${-2, 1}$ is a set whose sum of squares is unchanged by this process. – eyeballfrog Jan 27 '18 at 01:18
  • If each of fifty $-1$s and fifty $0$s is increased by $1$, the sum of the squares won’t change. Of course, that doesn’t mean there aren’t other possibilities for the $100$ numbers. – Steve Kass Jan 27 '18 at 01:47

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Hint:   in the case of $\,3\,$ numbers, the premise is: $$(a+1)^2+(b+1)^2+(c+1)^2 - (a^2+b^2+c^2) = 0 \;\;\iff\;\; 2(a+b+c)=-3$$

Then $\,(a+2)^2+(b+2)^2+(c+2)^2 - \big(a^2+b^2+c^2\big) = 4(a+b+c)+12=6\,$.

dxiv
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  • So the conclusion is that the sum will not change, but I am not sure how this applies to 100 numbers where 100 isn't divisible by 3. – Gerard L. Jan 27 '18 at 01:37
  • conclusion is that the sum will not change On the contrary, the sum will change since the RHS of the last line is not $,0,$. – dxiv Jan 27 '18 at 01:39