Checking whether $5876432$ is a perfect square or not I was solving a question and ended up with the expression $\sqrt{5809}$ and seeing this expression started a question in my head:If numbers upto $700$ are given to me i can immediately see them and tell whether they are perfect squares or not;but what if really really large numbers like 5876432 are given?
so,my question is-is there any method to know whether a large number like 5876432 is a perfect square or not(of course without using the calculator)?
Because knowing that allows me to use the prime factorisation method instead of that long division method.
I want a clear and detailed answer because i'm still a beginner in mathematics.
 A: No square number ends in a 2, 3, 7 or 8. This is because when multiplying numbers, it is only the ones place of each number that contributes to the ones place of the final answer. 
Going through $\ 1^2, 2^2, 3^2, ...,9^2$ no value ends in a 2, 3, 7 or 8. Notice that $\ (10n+k)^2 = 100n^2 + 20nk + k^2 $ and therefore clearly any square larger than or equal to 10 will have a ones value equal to the ones value of $\ k^2$, and therefore cannot include 2, 3, 7 or 8
EDIT: By writing a quick program in c++ and by a similar method to above but looking at the last two digits, only numbers ending in the following digits may POSSIBLY be perfect squares. All other numbers MUST not be.
0, 1, 4, 9, 16, 21, 24
25,
29,
36,
41,
44,
49,
56,
61,
64,
69,
76,
81,
84,
89,
96,
A: If you know the square numbers up to 700 then you know that $24^2 = 576 < 587 < 625 = 25^2$ and thus $2400^2 < 5876432 < 2500^2$.
You can then observe that $(2400 + k)^2 = 2400^2 + 4800k + k^2$, so if $(2400 + k)^2 = 5876432$, then $4800k + k^2 = 116432$. We know that $k < 100$, so that's really dominated by the $4800k$ term, so we can do a quick (ish) long division and find $k$ is about $24$.
Then compute $2424^2 = 5760000 + 2\times 57600 + 576 = 5875776$. Moreover, $2425^2$ is at least $4800$ bigger, so $2424^2 < 5876432 < 2425^2$.
A: Well, as to this case, no square ends in a 2. 
