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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.

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    $\begingroup$ Sounds like a job for casting out nines. The residue mod $9$ is independent of the order of digits. $\endgroup$ – hardmath Jan 26 '18 at 4:12
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    $\begingroup$ $\sum_{k=1}^9 {k \cdot k} \equiv 6 (mod \ 9)$. That's it. $\endgroup$ – Bram28 Jan 26 '18 at 4:18
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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

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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

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  • $\begingroup$ How do you find out the divisors, when there are about $$\frac {45!}{9!.8!.7!.6!.5!.4!.3!.2!}$$ numbers to select from. You cannot judge the number of divisors Until and unless you have a specific number in hand. $\endgroup$ – Rohan Shinde Jan 26 '18 at 7:09
  • $\begingroup$ For example consider 12 and 21 which are permutations of digits 1 and 2. 12 has 6 divisors while 21 has only 4 divisors. $\endgroup$ – Rohan Shinde Jan 26 '18 at 7:18

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