I'm sure there is a more elegant way to prove this, but ...
Assertion: for all odd n > 3, the base 10 representation of n^2 has an even digit in the ten's column.
(In fact, even 1 and 9 contain even digits if you count leading zeros, which we don't).
Suppose that for some odd n>3, n^2 has an even digit in the ten's column, but (n+2)^2 contains only odd digits.
(n+2)^2 = n^2 + 4n + 4
Consider the pattern (for odd n) of the final digit of n, and the final two digits of n^2, and (4n+4).
The ten's digit will be represented by 'e' for even, or the letter 'o' for odd.
n n^2 (4n+4) sum (n^2+4n+4)
1 e1 e8 e9
3 e9 o6 e5
5 e5 e4 e9
7 e9 o2 e1
9 e1 e0 e1
Posted by Larry
on 2020-10-05 09:58:51