Sorry I'm not CU, but I'll try this question. I've done a variant of it before I think.
Let's call N = 100a + 10b + c
Then we know:
100a + 10b + c = 11m for some arbitrary integer m
m = a^2 + b^2 + c^2
100a + 10b + c = 11a^2 + 11b^2 + 11c^2.
Using divisibility rule of 11, we know:
Case 1:
a-b+c = 0
a+c = b
or Case 2:
a-b+c = 11
b = a+c-11
this splits the problem into two cases for us:
1. b = a +c where the first case for divisibility of 11 holds true.
100a + c + 10a + 10c = 11a^2 + 11c^2 + 11(a+c)^2
10a + c = 2a^2 + 2ac+ 2c^2
10a + c = 2a^2 + 2ac + 2c^2
Here, we're kind of stuck. Realize that we can factor out 2 from the right hand side, meaning that c must be even since a is also even.
we can write c = 2z, with z being either 0, 1, 2, 3, or 4, and substitute it in
we get:
10 a + 2z = 2a^2 + 4az + 8q^2
Dividing by 2 on both sides, this gives us:
5a + z = a^2 + 2az + 4z^2
We just bash out cases now, and eventually we find:
when we have z = 0, the case works out, with a being 5, and c = 0 b = 5. From this case, we get that N = 550. All other cases don't work(2,4,6,8)
I'll leave the rest of the problem to you; there's only one other case!
2. ?????
also edit: sorry for the bad "typography". Had to do this in a hurry
