How to prove that there are no three such digits (0-9) whose sum of squares is 172?

Question

Mathematics

Conditions of the problem: there is a three-digit natural number n, as well as a function K(n), which sums the squares of the digits of this number. Does there exist a three-digit number n whose value in the function K(n) is 172 (to be proved)?

Answer 1

Bust.
98x - doesn't work
97x - doesn't work
96x - doesn't work
95x - doesn't work
94x - doesn't work
93x - x is too big
88x - doesn't work
87x - doesn't work
86x - doesn't work
85x - x is too big
77x - does n't work
76x - x is already too big too big
66x - x and even more too big
172 is divisible by 4. Any square is divisible by 4 with a remainder of 0 or 1 - therefore we must have three even numbers.
The overhead is reduced.
88x - it doesn't work
86x - x is already too big
66x - x is even too big, it makes no sense to sort it out further.

Answer 2

So go through all the numbers from 100 to 999 and see. Or do you need an analytical proof?