Since the representation is decimal, let n^2=(100x+10a+b)^2, with 0<{a,b}<10, and some x.
By inspection, all that affects the final 2 digits of n^2 is the expression 20ab+b^2.
20ab is always even.
As we are only interested in odd digits, b^2 is in {1,9,25,49,81}. If b is {1,3}, there is nothing to carry, and if not, we are adding even to even. Hence the penultimate digit is always even, if the final digit is odd. The sole exception is if a=x=0, so that there is no penultimate digit.
So {1,3} are the only solutions to the given problem.
Edited on October 6, 2020, 5:16 am
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Posted by broll
on 2020-10-06 04:57:32 |
a slightly shorter method.
