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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.

Durgeshwar Ojha
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3 Answers3

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Well, as to this case, no square ends in a 2.

Tom Miller
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  • can i have the proof of this thing that no square ends in a 2? – Durgeshwar Ojha Oct 10 '17 at 09:52
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    All numbers are one of 0,1,...,9 (mod 10). Squaring each of these, we find that squares are all either 0,1,4,5,6 or 9 (mod 10), i.e. their last digit is one of those and cannot be 2,3,7 or 8. – Tom Miller Oct 10 '17 at 09:55
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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,

Kieren Pearson
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    thanks man,your answer was really helpful,but what if i give you 5876434?what will you say then? – Durgeshwar Ojha Oct 10 '17 at 10:14
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    Well 34 is not in the list of required last two digits, so its not a perfect square – Kieren Pearson Oct 10 '17 at 10:27
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    you found out this list by writing a program in c++?please tell me something about that program. – Durgeshwar Ojha Oct 10 '17 at 10:32
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    Similar to how I only have to check squares 1-9 so find what ones places perfect squares can have, I can check numbers 1-99 to find what tens and ones place perfect squares must have. It's a very simple program that loops through and calculates all the square numbers 1-99 and makes a list of the last two digits of each of them – Kieren Pearson Oct 10 '17 at 10:35
  • One can also use casting out nines; a square has to be 1,4,7, or 9. – Acccumulation Oct 10 '17 at 13:58
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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 $2424^2 < 5876432 < 2425^2$.

Ben Millwood
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