>>7973576
bruteforce prime factorization with a really big computer.

>estimate a quality

>>7973576

What number it ends in is a good place to start. It can certainly rule some things out.

By doing it in reverse. Either you find two numbers i and j whose squares encapsulate your large number or you find your large number exactly.

>>7973576
Why would you care about this?

>>7973576
Mentally indicate to someone to put the root into a calculator

>>7973576
>divisible by 2/8/32 but not 4/16/64
>divisible by 3/27 but not 9/81
>divisible by 5/10/1000 but not 25/100/10000
>divisible by 11 but not 121

>>7973576
By assuming it won't be. Given there are 899 squares between 10000 and 999999 your "estimate" will be correct in 99.9909% of cases

>>7973576
>estimate a binary outcome
This will surely work well.

>>7973576
The square of numbers 1-10: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

Tkae the last digit of each (ignoring repeats): 1, 4, 9, 6, 5, 0.

If your large number doesn't end in one of these digits, then it's definitely not a perfect square. That's the best I got at the moment, but this is an interesting question and I'll think on it.

>>7973756
This

>>7973914
This.

0% of numbers are squares.

>>7973576

OP's goofy question isn't about primes, but I found a very nice prime number range searcher here, which deserves sharing:

http://www.naturalnumbers.org/primes.html

It kicks out primes over a given range to text files, and works very nicely for the first million, the range I looked at.

But let's consider a version of OP's question. Let's consider all five to six digit natural numbers, and all the n, such that 10,000 <= n^2 <= 999,999.

Now, since 999,999 - 9,999 = 990,000, there are almost (but not quite) a million numbers in the above range. It wouldn't be unreasonable to include the first 9,999 in a consideration up to 1M, but let's stick with OP's thing and stay literal. We have the further bounds that 100^2 = 10,000 (in the range), and that 1,000^2 = 1,000,000, just outside its upper limit. So, every square from 100 up until 999 inclusive, or 900 numbers, are uniquely mapped to OP's spec in the above range of 990,000. 900/990,000 works out to a 1 in 1100 chance that some random number from the range is a perfect square, and this only spreads out more when working with like ranges over higher-number intervals, so to answer OP's question, your best bet is to "estimate" that a given number is not a perfect square.

However, this can be refined further by making an observation about the ending digits of perfect squares: 0,1,4,9,6,5,6,9,4,1,0,1,4,9,6.... Proving that this pattern always holds is left as an exercise, but when confronted with a natural number ending in 2,3,7 or 8, per the above it can be immediately ruled out as a perfect square (another exercise). Thus you can instantly rule out 40% or thereabouts of a given range as a potential perfect square, absent other information.

Although there are power functions that grow more rapidly of course, when seen in this prosaic light, perfect squares are rather rare items, themselves.

>>7974491
50% are squares

>>7973576
It's last digit will be the same as the last digit in one of the following numbers.
0,1,4,9,16,25,36,49,64,81.
Pretty much, if it ends in 2,3,7 or 8, it won't have a nice square root.

>>7974887
You might be making a joke, but let #(n) denote the number of squares less than n.

The limit as n approaches infinity of #(n)/n, the proportion of numbers that are squares, is 0.

>>7976355
Not in Finite Fields.

>>7976371
True, and that's relevant how?

>>7976355
I was kind of making a joke, but also not. The percentage of natural numbers that are perfect squares is zero, but there are as many perfect squares as there are natural numbers.