Let s(N) denote the sum of the digits of N. Find the sum of all two-digit numbers N for which s(N^2)=s(N)^2.
sum(N^2) = N(N + 1)(2N + 1)/6
= 1 + 4 + 9 + 16 + ...
by s(N)^2 , I assume you mean [sum(N)]^2, or else it is the same as the first
sum(n) = N(N+1)/2
[sum(N)]^2 = 1^2 + 3^2 + 6^2 + 10^2 + 15^2 + 21^2 + ...
Wolfram found that
1^2 + 3^2 + 6^2 + 10^2 + 15^2 + 21^2 + ... = sigma (1/4)N^2(1+N)^2
https://www.wolframalpha.com/input/?i=pattern+%7B1%5E2+%2B+3%5E2+%2B+6%5E2+%2B+10%5E2+%2B+15%5E2+%2B+21%5E2+%2B+...%7D
(1/4)N^2(1+N)^2 = N(N + 1)(2N + 1)/6
divide by N(N+1)
(1/4)N(1+N) = (2N+1)/6
Times 12
3N(1+N) = 2(2N+1), switching to n
3n + 3n^2 = 4n + 4
3n^2 - n - 4 = 0
(n-1)(3n+4) = 0
n =1 or n is a fraction, but n must be a whole number
test:
if n = 1, sum(1^2) = 1
[sum(1)]^2 = 1 , works
if n = 2, sum(2^2) = 1 + 4 = 5
[sum(2)]^2 = 1^2 + 3^2 = 10 , no good
in n = 3, sum(3^2) = 1 + 4 + 9 = 14
[sum(3)]^2 = 1^2 + 3^2 + 6^2 = 46
clearly [sum(N)]^2 diverges faster and they can never be the same again
N =1
Did I overthink the question???