Prove that any permutation of the given numbers can't be a perfect square

Question

Prove that a natural number written using one $1$, two $2$'s, three $3$'s, ... , nine $9$'s cannot be a perfect square.

This is the first problem that I have encountered of any such types. So I was not getting even a single idea over how to start. I don't have a proper understanding of the mathematical induction so any other method or a hint will be really appreciable.

Answer 1

Using the hint as given by Bram28 the brief explanation to the hint is:-

The sum of digits of every perfect square follows exactly one condition from the below given four conditions. Let $n^2$ be any perfect square and $k$ represent the Sum of the digits of $n^2$ then

Either $k\equiv 0\pmod 9$

Or $k\equiv 1\pmod 9$

Or $k\equiv 4\pmod 9$

Or $k\equiv 7\pmod 9$

Here in this case the sum of digits of the number formed by any permutation of one $1$, two $2$'s, three $3$'s,.... , nine $9$'s would be $$\sum_{i=1}^9 i^2=285$$

Now $$285\equiv 6\pmod 9$$

Hence any permutation of given digits won't form a perfect square.

Q. E. D

Answer 2

I dont know a solution but maybe a hint that might be helpful: A property of square numbers is that they are the only whole numbers divisible by a odd number of whole numbers . These numbers you have to prove they are not squared, try to look at divisors