I was going through the fundamental theorem in Number Theory where any non zero integer n can be represented as a product of distinct primes. A related problem with this theorem is to prove that for every such number, there exists a prime $p$ such that $p< \sqrt n$.
I was wondering if there is any mathematical proof that no prime $p$ exists for the number $n$ such that $p> \sqrt n$.

  • $\begingroup$ ... but if you move $1/2$ down from the exponent in $n^{1/2}$, it's correct: $p\le \frac{1}{2}n$ $\endgroup$
    – draks ...
    Commented Jun 21, 2012 at 6:18
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    $\begingroup$ ¿did you see at a simple example? 15=3 * 5 and 5 is a prime certainly larget than the square root of 15! $\endgroup$ Commented Sep 28, 2012 at 0:24
  • $\begingroup$ proofwiki.org/wiki/… $\endgroup$
    – Trancot
    Commented Apr 29, 2013 at 6:08
  • $\begingroup$ Wouldn'd this imply that the set of all prime numbers would be finite, limited by the square root of any number??? $\endgroup$ Commented Jun 1, 2016 at 11:08
  • $\begingroup$ @EddeAlmeida: No, because this is only talking about the prime factors of one specific n, not of all integers. $\endgroup$
    – smci
    Commented Jan 23, 2017 at 9:53

8 Answers 8


No. Consider that the square root of $14$ is about $3.74$ but $14$ has $7$ as a prime factor. Also consider that any prime number such as $2$ is its own (only) prime factor, and any number greater than $1$ is greater than its square root. The theorem you have stated is incorrect: $25$ has no prime factor less than $5$, and $3$ has no prime factor less than $1.732$; however, it is true that every composite number has a prime factor less than or equal to its square root.

  • 1
    $\begingroup$ Ok, now I can see it. If such a prime p exists such that $p$ > $\sqrt n$ and it is not equal to n then it must have a prime $q$ < $\sqrt n$. Otherwise all such primes should be less than $\sqrt n$. Thus I get the solution to my related problem. Thnx for help. $\endgroup$ Commented Sep 10, 2011 at 8:24
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    $\begingroup$ That is true, unless n is a square of a prime (e.g. 25), in which case its prime factor is exactly equal to its square root. Now what if n is the square of a composite (e.g. 36)? $\endgroup$ Commented Sep 10, 2011 at 8:39
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    $\begingroup$ missed out mentioning n was composite...my bad!! Was more concerned about how to represent $\sqrt n$ here :) $\endgroup$ Commented Sep 10, 2011 at 9:22
  • $\begingroup$ But then while doing prime factorization number it is said check primes only till root of the number. Why? Where I am getting confused? $\endgroup$ Commented Sep 22, 2017 at 6:03
  • $\begingroup$ Is it true that at most there exists only one prime factor strictly greater than the square root and that it would be of order one? All the counterexamples along the lines of this answer that I can find (6, 10, 15, 20, 21) all have only one factor above the root. $\endgroup$
    – Double AA
    Commented Jan 11, 2018 at 18:43

You seem to be confused with another statement, which is that the smallest prime factor of a composite number N is less than or equal to $\sqrt N$.

  • $\begingroup$ No actually the question which got me thinking the above asked question was--> Show that there exists a prime p dividing $n$ (composite number), with $p$ <= $\sqrt n$ . $\endgroup$ Commented Sep 10, 2011 at 8:48
  • $\begingroup$ @AbhishekAnand The statements "the smallest prime factor of a composite number$N$ is less than or equal to $\sqrt N$" (in Mark Bennet's answer) and "there exists a prime $p$ dividing $n$ (composite number) with $p\leq\sqrt n$" (in your comment) are obviously equivalent. If some prime factor is $\leq\sqrt n$, then the smallest prime factor can't be any bigger. $\endgroup$ Commented Jul 31, 2014 at 1:04
  • $\begingroup$ Hey Mark, don't you mean the "largest" prime factor of a composite number... I mean wouldn't the smallest prime factor of every composite number be 1? Or am I missing something? $\endgroup$ Commented Oct 19, 2019 at 16:28
  • $\begingroup$ @TejasShah There are some people who count $1$ as a prime. I don't - it is best considered as a unit rather than a prime. $\endgroup$ Commented Oct 19, 2019 at 20:32

There can be at most $1$ prime factor of $n$ greater than $\sqrt{n}$.

Proof: If possible let there be two distinct prime factors of $n$ which are greater than $\sqrt{n}$, say $a$ and $b$. As both $a$ and $b$ are different prime factors of $n$ their product $c = a \times b$ should also be the factor of $n$. But $c$ would exceed $n$ as both $a$ and $b$ are greater than $\sqrt{n}$, which contradicts our assumption, hence there can only be one prime factor of $n$ greater than $\sqrt{n}$.

Application: If u want to do prime factorization of $n$ then you only need primes up to $\sqrt{n}$ which largely improves the complexity of prime factorization. The number left after repeatedly dividing $n$ with all primes factors less than $\sqrt{n}$ is indeed a prime or $1$.

Repeatedly dividing $n$ with all primes factors less than $\sqrt{n}$:

You iterate over all primes less than $\sqrt{n}$. If the current prime $p$ divides $n$ you keep on dividing $n$ by $p$ until $n$ becomes $1$ or the $p$ no longer divides $n$. In the end, either $n$ will become $1$ or some number greater than $\sqrt{n}$ would be left which would also be a prime factor of $n$ but it will be the only prime factor which is greater than $\sqrt{n}$


Proof: Suppose $n$ is a positive integer s.t. $n=pq$, where $p$ and $q$ are prime numbers. Assume $p>\sqrt{n}$ and $q>\sqrt{n}$. Multiplying these inequalities we have $p.q>\sqrt{n}.\sqrt{n}$, which implies $pq>n$. This is a contradiction to our hypothesis $n=pq$. Hence we can conclude that either $p\leq \sqrt{n}$ or $q \leq \sqrt{n}$.

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    $\begingroup$ Welcome to MathSE! Your proof shows if $n$ is the product of two primes, then at most one (prime) factor is greater than $\sqrt{n}$ but this late answer doesn't contribute much to what others said and doesn't directly address the Question, is every prime factor of a positive integer less than its square root? $\endgroup$
    – hardmath
    Commented Jun 21, 2012 at 14:35
  • $\begingroup$ proofwiki.org/wiki/… $\endgroup$
    – Trancot
    Commented Apr 29, 2013 at 6:08

It doesn't mean that every factor of $n$ would be less that $\sqrt{n}$, in fact at least one factor would be less than $\sqrt{n}$ if $n$ is not a prime number. Explanation: $$ n=\sqrt{n}\cdot \sqrt{n},\quad n=a\cdot b, $$ so 1) if one factor is less than $\sqrt{n}$ then other will be greater than $\sqrt{n}$, 2) if there is no such factor less than $\sqrt{n}$ then both factors would be greater than $\sqrt{n}$ but it's not possible; so, that number must be prime if it doesn't have a factor less than $\sqrt{n}$.


I know the answer is late. But, maybe its useful for reference.

For a number n consider these cases:

  1. The given number is prime and has only itself and 1 as factors. (1,n)
  2. The given number is a square of a prime p
        => n=p^2 or n=p*p
  3. The given number is a product of just two prime numbers p & q
        => n=p*q
  4. The given number has again two prime factors, but now, and at-least one of them is repeated
        => n = p*p*q OR p*p*p*q OR p*q*q ... etc.
  5. The given number has more than two prime factors, each with/without repetition
        => n = p*q*r*... etc.

1 - No other prime factors -- Eg. 11
2 - p=sqrt(n) -- Eg. 5=sqrt(25)
3 - one of p,q will be "< sqrt(n)" and the other will be "> sqrt(n)" -- Eg. 2*7=14
4 - At most one out of p & q could be "> sqrt(n)" -- Eg. 2*2*7=28
5 - At most one of them could be "> sqrt(n)" -- Eg. 2*2*3*7=84

Clearly, the only cases 1,2,3 touch or cross the sqrt(n) border.
Case 1 Is crossing because its only factor is itself
Case 2 Is touching sqrt.. cant cross
Case 3 Will cross sqrt because fac1<sqrt and fac2>sqrt
Case 4 & 5 Could cross sqrt sometimes

Think of square-root as splitting a number into two in the multiplicative sense.
Case 1 - you cant split
Case 2 - your doing the same thing.. splitting into two
Case 3 - you can't split perfectly as the number is not a square. So, there is an unequal split with one side dominating
Case 4 - you are splitting more than twice. The pieces become way too small. But consider this case: if p*p is itself <sqrt(n) q could still be >sqrt(n)
Case 5 - again you are splitting more than twice. The pieces become way too small. But one piece could still become greater than sqrt like in case 4 above

If n is not a perfect square, factors can be bunched up so that one chain will be <sqrt(n) and the other chain will be >sqrt(n). If the chain that is >sqrt(n) has only one member, in that case the prime-factor that is >sqrt(n) exists.

  • $\begingroup$ Your statement is confusing. Does $42$ comply with this or not? $\endgroup$
    – WimC
    Commented Apr 28, 2013 at 7:19
  • $\begingroup$ @WimC I had a lot of corrections to make. Does it make sense now? $\endgroup$
    – Theo
    Commented Apr 28, 2013 at 8:30

N=a*b=ab+b^2-b^2=b^2 + b(a-b)

if b < a then b^2 < N so b < sqrt(N)

if b = a a = b = sqrt(N)


A prime factor is not always less than the square root as you could see with the $14 = 2 \times 7$ as a counter example, since $7 \gt \sqrt14$. But you can show that when the number is not prime, the smallest prime divisor less than the number itself is less than or equal to the square root.

Let $n$ be a positive composite integer and $d$ be the smallest prime factor of $n$ less than $n$ itself.

Write $n = dq$ for some integer $q \gt 0$. We should have $q \gt 1$, otherwise we would have $n = d$. So, either $q \lt d$ or $q \ge d$.

Suppose then that $q \lt d$. Then either $q$ is prime or a product of primes. In either cases we would have found prime divisors of $n$ less than $d$, a contradiction to the fact that $d$ is the smallest. So $q \ge d$, and we can divide $q$ by $d$ and write $q = kd + r$ for some integer $k$ and $0 \le r \lt d$. Substituting this back in the first equation you get $n = d(kd + r) = d^2k + dr \ge d^2$, therefore $d^2 \le n$ and so $d \le \sqrt n$.

So clearly the smallest prime factor of a number is either the number itself in case it's prime or it's less than or equal to the square root of the number. An upper bound for divisors less than $n$ would be $\frac{n}{2}$. You can look at $n = dq$ and see that you want $q >= 2$ again and in this case $d$ is maximum when $q = 2$ and you would have $n = 2d$ so $d = \frac{n}{2}$.


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