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Please explain how you can do this?

helloworld
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    There's a bunch of properties you can use to rule numbers out. For instance, if it ends in a $0$, it must end in an even number of $0's$, otherwise you won't have the right paired numbers of factors of 2 and 5. – Alan Mar 19 '15 at 01:07
  • Following @Alan's suggestion, try to find the prime divisors if possible and factorize the numbers. If it is a perfect square, then the largest prime could be at most the squareroot of the number, so you may stop your search for primes once they are half of the number of digits. Finding the prime factorization however can be rather difficult to do by hand. – JMoravitz Mar 19 '15 at 01:11
  • Read here for some common divisibility tests for numbers under 50. – JMoravitz Mar 19 '15 at 01:12
  • I've figured it out using properties of perfect squares. Thanks for the help guys. – helloworld Mar 19 '15 at 01:25
  • If you have figured it out yourself, using properties of perfect squares, can you share your insights with others by posting your own answer? – Greenonline Mar 19 '15 at 02:33

3 Answers3

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This is not a complete answer, but I doubt that there is a complete answer. For example I would not expect any easy way to tell whether or not $990297961$ is a square.

However . . . I would start by looking at the remainders when your numbers are reduced modulo $8,9$ and $5$, and seeing whether they match the possible squares in those moduli. $$\matrix{ &\hbox{mod $8$}&\hbox{mod $9$}&\hbox{mod $5$}\cr \hbox{possible squares}&0,1,4&0,1,4,7&0,1,4\cr 23921881&1&7&1\cr 5312660544&0&0&4\cr 445382688&0&\color{red}{3}\cr 1407525289&\color{red}{3}\cr}$$ The red entries above show that your third and fourth examples are not squares; the first two are still unknown. You could try doing something for further moduli, perhaps $11$ and $7$. But while this may rule out some numbers, it won't (at least not easily) show you that something is a square.

David
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There are ways to establish that a number is NOT a square. If the number is divisible by 2 but not by 4, or divisible by 3 but not by 9, or divisible by 5 but not by 25, those cannot be squares. A square must end in one of {0,1,4,5,6,9}. There are only 22 choices for the last digit pair: $$\{00,01,04,09,16,21,24,25,29,36,41,44,49,56,61,64,69,76,81,84,89,96\}$$

But I don't know of an easy-to-apply test that will definitely say that an arbitrary large number is a square.

Joffan
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Determining which numbers are perfect squares by hand is a rather tedious and difficult process which is often equivalent to finding what the prime factorization of the number is. As a result, it might require a huge amount of patience, or might be completely infeasible to do by hand for large numbers.

Without completing the factorization, you cannot confirm a number is a perfect square, but if you notice something wrong during the factorization process you can show it is not a perfect square before completing the process.

Direct your attention to the number $445382688$. For our test of divisibility, we see that the number is divisible by two (since the last digit is even). If it is in fact a perfect square, then it should be divisible by an even power of two. We check if it is divisible by four by looking at the final two digits, $88$. $88$ is in fact a multiple of four ($88=22\cdot 4$) so we can divide by four and look at the result.

$445382688/4=111345672$. We see the new number is again even, so we run the test for divisibility by four. The final two digits, $72$, is again a multiple of four ($72=18\cdot 4$), so we divide by four again.

$111345672/4=27836418$. We see this new number is yet again even, so we run the test for divisibility by four one more time. The final two digits, $18$ however is not divisible by four, so we see that in the prime factorization of the original number, two appears five times. Therefore it is not a perfect square.

Using this number as an example again, and ignoring the fact that 2 has an odd number of powers, you could see it is not a perfect square using divisibility by 3's and 9's.

$445382688$: To test for divisibility by three, add the digits together. $445382688\Rightarrow 4+4+5+3+8+2+6+8+8 = 48$. If you don't immediately recognize if the result is a multiple of three, you may repeat this process. $48\Rightarrow 4+8 = 12\Rightarrow 1+2 = 3$. If the result (after any number of iterations) is divisible by three, then the original number is too.

Divisibility by nine works very similarly, however you require that the result after repeated summation of digits is exactly nine (or a multiple of 9). $3$ is not a multiple of $9$, so the original number 445382688 is divisible by 3 but not divisible by 9, so again we see that it is not a perfect square.

JMoravitz
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