Question

Prove that: `(s-a_1)^2 + (s-a_2)^2 +cdots+ (s-a_n)^2 =a_1^2 +a_2^2+cdots+a_n^2` when `a_1^2 +a_2^2+cdots+a_n^2 = n/{2s}`?

Answer 1

The given identity is false, but if:

`a_1+a_2+...+a_n = n/2 s`

then:

`(s-a_1)^2+(s-a_2)^2+...+(s-a_n)^2 = a_1^2+a_2^2+...+a_n^2`

Explanation:

Consider the case `n=1`, `s=2`, `a_1 = 1/2`

Then:

`a_1^2 = (1/2)^2 = 1/4 = 1/(2*2) = n/(2s)`

`(s-a_1)^2 = (2-1/2)^2 = (3/2)^2 = 9/4 != 1/4 = a_1^2`

So something is not quite right with the question as stands.

Bonus

Let's try to wo##rk out the correct condition.

Working back from the desired result:

`(s-a_1)^2+(s-a_2)^2+...+(s-a_n)^2 = a_1^2+a_2^2+...+a_n^2`

Multiplying out the left hand side, we have:

`a_1^2+a_2^2+...+a_n^2-2s(a_1+a_2+...+a_n)+ns^2 = a_1^2+a_2^2+...+a_n^2`

Subtract `a_1^2+a_2^2+...+a_n^2` from both sides to get:

`-2s(a_1+a_2+...+a_n)+ns^2 = 0`

Add `2s(a_1+a_2+...+a_n)` to both sides to get:

`ns^2 = 2s(a_1+a_2+...+a_n)`

Divide both sides by `2s` to get:

`n/2 s = a_1+a_2+...+a_n`

Note that all of the above steps are reversible, so if:

`a_1+a_2+...+a_n = n/2 s`

then:

`(s-a_1)^2+(s-a_2)^2+...+(s-a_n)^2 = a_1^2+a_2^2+...+a_n^2`