Question

what is the smallest of three consecutive numbers when squared add up to 365

Answer 1

If the smallest is x, then

x^2 + (x+1)^2 + (x+2)^2 = 365
x=10

Or, since the three numbers are consecutive, they are all almost equal. So, if the middle number is x, then
3x^2 ≈ 365
x^2 ≈ 121
x = 11
That makes the smallest equal to 10.