Check if a number is a perfect square


How could I check if a number is a perfect square?

Speed is of no concern, for now, just working.

See also: Integer square root in python.

Asked By: delete



You could binary-search for the rounded square root. Square the result to see if it matches the original value.

You’re probably better off with FogleBirds answer – though beware, as floating point arithmetic is approximate, which can throw this approach off. You could in principle get a false positive from a large integer which is one more than a perfect square, for instance, due to lost precision.

Answered By: user180247

The problem with relying on any floating point computation (math.sqrt(x), or x**0.5) is that you can’t really be sure it’s exact (for sufficiently large integers x, it won’t be, and might even overflow). Fortunately (if one’s in no hurry;-) there are many pure integer approaches, such as the following…:

def is_square(apositiveint):
  x = apositiveint // 2
  seen = set([x])
  while x * x != apositiveint:
    x = (x + (apositiveint // x)) // 2
    if x in seen: return False
  return True

for i in range(110, 130):
   print i, is_square(i)

Hint: it’s based on the “Babylonian algorithm” for square root, see wikipedia. It does work for any positive number for which you have enough memory for the computation to proceed to completion;-).

Edit: let’s see an example…

x = 12345678987654321234567 ** 2

for i in range(x, x+2):
   print i, is_square(i)

this prints, as desired (and in a reasonable amount of time, too;-):

152415789666209426002111556165263283035677489 True
152415789666209426002111556165263283035677490 False

Please, before you propose solutions based on floating point intermediate results, make sure they work correctly on this simple example — it’s not that hard (you just need a few extra checks in case the sqrt computed is a little off), just takes a bit of care.

And then try with x**7 and find clever way to work around the problem you’ll get,

OverflowError: long int too large to convert to float

you’ll have to get more and more clever as the numbers keep growing, of course.

If I was in a hurry, of course, I’d use gmpy — but then, I’m clearly biased;-).

>>> import gmpy
>>> gmpy.is_square(x**7)
>>> gmpy.is_square(x**7 + 1)

Yeah, I know, that’s just so easy it feels like cheating (a bit the way I feel towards Python in general;-) — no cleverness at all, just perfect directness and simplicity (and, in the case of gmpy, sheer speed;-)…

Answered By: Alex Martelli

Use Newton’s method to quickly zero in on the nearest integer square root, then square it and see if it’s your number. See isqrt.

Python ≥ 3.8 has math.isqrt. If using an older version of Python, look for the “def isqrt(n)” implementation here.

import math

def is_square(i: int) -> bool:
    return i == math.isqrt(i) ** 2

Since you can never depend on exact comparisons when dealing with floating point computations (such as these ways of calculating the square root), a less error-prone implementation would be

import math

def is_square(integer):
    root = math.sqrt(integer)
    return integer == int(root + 0.5) ** 2

Imagine integer is 9. math.sqrt(9) could be 3.0, but it could also be something like 2.99999 or 3.00001, so squaring the result right off isn’t reliable. Knowing that int takes the floor value, increasing the float value by 0.5 first means we’ll get the value we’re looking for if we’re in a range where float still has a fine enough resolution to represent numbers near the one for which we are looking.

Answered By: Mike Graham
  1. Decide how long the number will be.
  2. take a delta 0.000000000000…….000001
  3. see if the (sqrt(x))^2 – x is greater / equal /smaller than delta and decide based on the delta error.

This response doesn’t pertain to your stated question, but to an implicit question I see in the code you posted, ie, “how to check if something is an integer?”

The first answer you’ll generally get to that question is “Don’t!” And it’s true that in Python, typechecking is usually not the right thing to do.

For those rare exceptions, though, instead of looking for a decimal point in the string representation of the number, the thing to do is use the isinstance function:

>>> isinstance(5,int)
>>> isinstance(5.0,int)

Of course this applies to the variable rather than a value. If I wanted to determine whether the value was an integer, I’d do this:

>>> x=5.0
>>> round(x) == x

But as everyone else has covered in detail, there are floating-point issues to be considered in most non-toy examples of this kind of thing.

Answered By: Vicki Laidler

I just posted a slight variation on some of the examples above on another thread (Finding perfect squares) and thought I’d include a slight variation of what I posted there here (using nsqrt as a temporary variable), in case it’s of interest / use:

import math

def is_square(n):
  if not (isinstance(n, int) and (n >= 0)):
    return False 
    nsqrt = math.sqrt(n)
    return nsqrt == math.trunc(nsqrt)

It is incorrect for a large non-square such as 152415789666209426002111556165263283035677490.

Answered By: gumption
import math

def is_square(n):
    sqrt = math.sqrt(n)
    return (sqrt - int(sqrt)) == 0

A perfect square is a number that can be expressed as the product of two equal integers. math.sqrt(number) return a float. int(math.sqrt(number)) casts the outcome to int.

If the square root is an integer, like 3, for example, then math.sqrt(number) - int(math.sqrt(number)) will be 0, and the if statement will be False. If the square root was a real number like 3.2, then it will be True and print “it’s not a perfect square”.

It fails for a large non-square such as 152415789666209426002111556165263283035677490.

Answered By: 0xPwn

This can be solved using the decimal module to get arbitrary precision square roots and easy checks for “exactness”:

import math
from decimal import localcontext, Context, Inexact

def is_perfect_square(x):
    # If you want to allow negative squares, then set x = abs(x) instead
    if x < 0:
        return False

    # Create localized, default context so flags and traps unset
    with localcontext(Context()) as ctx:
        # Set a precision sufficient to represent x exactly; `x or 1` avoids
        # math domain error for log10 when x is 0
        ctx.prec = math.ceil(math.log10(x or 1)) + 1  # Wrap ceil call in int() on Py2
        # Compute integer square root; don't even store result, just setting flags
        # If previous line couldn't represent square root as exact int, sets Inexact flag
        return not ctx.flags[Inexact]

For demonstration with truly huge values:

# I just kept mashing the numpad for awhile :-)
>>> base = 100009991439393999999393939398348438492389402490289028439083249803434098349083490340934903498034098390834980349083490384903843908309390282930823940230932490340983098349032098324908324098339779438974879480379380439748093874970843479280329708324970832497804329783429874329873429870234987234978034297804329782349783249873249870234987034298703249780349783497832497823497823497803429780324
>>> sqr = base ** 2
>>> sqr ** 0.5  # Too large to use floating point math
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
OverflowError: int too large to convert to float

>>> is_perfect_power(sqr)
>>> is_perfect_power(sqr-1)
>>> is_perfect_power(sqr+1)

If you increase the size of the value being tested, this eventually gets rather slow (takes close to a second for a 200,000 bit square), but for more moderate numbers (say, 20,000 bits), it’s still faster than a human would notice for individual values (~33 ms on my machine). But since speed wasn’t your primary concern, this is a good way to do it with Python’s standard libraries.

Of course, it would be much faster to use gmpy2 and just test gmpy2.mpz(x).is_square(), but if third party packages aren’t your thing, the above works quite well.

Answered By: ShadowRanger

If you want to loop over a range and do something for every number that is NOT a perfect square, you could do something like this:

def non_squares(upper):
    next_square = 0
    diff = 1
    for i in range(0, upper):
        if i == next_square:
            next_square += diff
            diff += 2
        yield i

If you want to do something for every number that IS a perfect square, the generator is even easier:

(n * n for n in range(upper))
Answered By: Moberg
import math

def is_square(n):
    sqrt = math.sqrt(n)
    return sqrt == int(sqrt)

It fails for a large non-square such as 152415789666209426002111556165263283035677490.

Answered By: Ravi Sankar Raju

This is my method:

def is_square(n) -> bool:
    return int(n**0.5)**2 == int(n)

Take square root of number. Convert to integer. Take the square. If the numbers are equal, then it is a perfect square otherwise not.

It is incorrect for a large square such as 152415789666209426002111556165263283035677489.

Answered By: Archu Sm

If youre interested, I have a pure-math response to a similar question at math stackexchange, “Detecting perfect squares faster than by extracting square root”.

My own implementation of isSquare(n) may not be the best, but I like it. Took me several months of study in math theory, digital computation and python programming, comparing myself to other contributors, etc., to really click with this method. I like its simplicity and efficiency though. I havent seen better. Tell me what you think.

def isSquare(n):
    ## Trivial checks
    if type(n) != int:  ## integer
        return False
    if n < 0:      ## positivity
        return False
    if n == 0:      ## 0 pass
        return True

    ## Reduction by powers of 4 with bit-logic
    while n&3 == 0:    

    ## Simple bit-logic test. All perfect squares, in binary,
    ## end in 001, when powers of 4 are factored out.
    if n&7 != 1:
        return False

    if n==1:
        return True  ## is power of 4, or even power of 2

    ## Simple modulo equivalency test
    c = n%10
    if c in {3, 7}:
        return False  ## Not 1,4,5,6,9 in mod 10
    if n % 7 in {3, 5, 6}:
        return False  ## Not 1,2,4 mod 7
    if n % 9 in {2,3,5,6,8}:
        return False  
    if n % 13 in {2,5,6,7,8,11}:
        return False  

    ## Other patterns
    if c == 5:  ## if it ends in a 5
        if (n//10)%10 != 2:
            return False    ## then it must end in 25
        if (n//100)%10 not in {0,2,6}: 
            return False    ## and in 025, 225, or 625
        if (n//100)%10 == 6:
            if (n//1000)%10 not in {0,5}:
                return False    ## that is, 0625 or 5625
        if (n//10)%4 != 0:
            return False    ## (4k)*10 + (1,9)

    ## Babylonian Algorithm. Finding the integer square root.
    ## Root extraction.
    s = (len(str(n))-1) // 2
    x = (10**s) * 4

    A = {x, n}
    while x * x != n:
        x = (x + (n // x)) >> 1
        if x in A:
            return False
    return True

Pretty straight forward. First it checks that we have an integer, and a positive one at that. Otherwise there is no point. It lets 0 slip through as True (necessary or else next block is infinite loop).

The next block of code systematically removes powers of 4 in a very fast sub-algorithm using bit shift and bit logic operations. We ultimately are not finding the isSquare of our original n but of a k<n that has been scaled down by powers of 4, if possible. This reduces the size of the number we are working with and really speeds up the Babylonian method, but also makes other checks faster too.

The third block of code performs a simple Boolean bit-logic test. The least significant three digits, in binary, of any perfect square are 001. Always. Save for leading zeros resulting from powers of 4, anyway, which has already been accounted for. If it fails the test, you immediately know it isnt a square. If it passes, you cant be sure.

Also, if we end up with a 1 for a test value then the test number was originally a power of 4, including perhaps 1 itself.

Like the third block, the fourth tests the ones-place value in decimal using simple modulus operator, and tends to catch values that slip through the previous test. Also a mod 7, mod 8, mod 9, and mod 13 test.

The fifth block of code checks for some of the well-known perfect square patterns. Numbers ending in 1 or 9 are preceded by a multiple of four. And numbers ending in 5 must end in 5625, 0625, 225, or 025. I had included others but realized they were redundant or never actually used.

Lastly, the sixth block of code resembles very much what the top answerer – Alex Martelli – answer is. Basically finds the square root using the ancient Babylonian algorithm, but restricting it to integer values while ignoring floating point. Done both for speed and extending the magnitudes of values that are testable. I used sets instead of lists because it takes far less time, I used bit shifts instead of division by two, and I smartly chose an initial start value much more efficiently.

By the way, I did test Alex Martelli’s recommended test number, as well as a few numbers many orders magnitude larger, such as:

x=1000199838770766116385386300483414671297203029840113913153824086810909168246772838680374612768821282446322068401699727842499994541063844393713189701844134801239504543830737724442006577672181059194558045164589783791764790043104263404683317158624270845302200548606715007310112016456397357027095564872551184907513312382763025454118825703090010401842892088063527451562032322039937924274426211671442740679624285180817682659081248396873230975882215128049713559849427311798959652681930663843994067353808298002406164092996533923220683447265882968239141724624870704231013642255563984374257471112743917655991279898690480703935007493906644744151022265929975993911186879561257100479593516979735117799410600147341193819147290056586421994333004992422258618475766549646258761885662783430625 ** 2
for i in range(x, x+2):
    print(i, isSquare(i))

printed the following results:

1000399717477066534083185452789672211951514938424998708930175541558932213310056978758103599452364409903384901149641614494249195605016959576235097480592396214296565598519295693079257885246632306201885850365687426564365813280963724310434494316592041592681626416195491751015907716210235352495422858432792668507052756279908951163972960239286719854867504108121432187033786444937064356645218196398775923710931242852937602515835035177768967470757847368349565128635934683294155947532322786360581473152034468071184081729335560769488880138928479829695277968766082973795720937033019047838250608170693879209655321034310764422462828792636246742456408134706264621790736361118589122797268261542115823201538743148116654378511916000714911467547209475246784887830649309238110794938892491396597873160778553131774466638923135932135417900066903068192088883207721545109720968467560224268563643820599665232314256575428214983451466488658896488012211237139254674708538347237589290497713613898546363590044902791724541048198769085430459186735166233549186115282574626012296888817453914112423361525305960060329430234696000121420787598967383958525670258016851764034555105019265380321048686563527396844220047826436035333266263375049097675787975100014823583097518824871586828195368306649956481108708929669583308777347960115138098217676704862934389659753628861667169905594181756523762369645897154232744410732552956489694024357481100742138381514396851789639339362228442689184910464071202445106084939268067445115601375050153663645294106475257440167535462278022649865332161044187890625 True
1000399717477066534083185452789672211951514938424998708930175541558932213310056978758103599452364409903384901149641614494249195605016959576235097480592396214296565598519295693079257885246632306201885850365687426564365813280963724310434494316592041592681626416195491751015907716210235352495422858432792668507052756279908951163972960239286719854867504108121432187033786444937064356645218196398775923710931242852937602515835035177768967470757847368349565128635934683294155947532322786360581473152034468071184081729335560769488880138928479829695277968766082973795720937033019047838250608170693879209655321034310764422462828792636246742456408134706264621790736361118589122797268261542115823201538743148116654378511916000714911467547209475246784887830649309238110794938892491396597873160778553131774466638923135932135417900066903068192088883207721545109720968467560224268563643820599665232314256575428214983451466488658896488012211237139254674708538347237589290497713613898546363590044902791724541048198769085430459186735166233549186115282574626012296888817453914112423361525305960060329430234696000121420787598967383958525670258016851764034555105019265380321048686563527396844220047826436035333266263375049097675787975100014823583097518824871586828195368306649956481108708929669583308777347960115138098217676704862934389659753628861667169905594181756523762369645897154232744410732552956489694024357481100742138381514396851789639339362228442689184910464071202445106084939268067445115601375050153663645294106475257440167535462278022649865332161044187890626 False

And it did this in 0.33 seconds.

In my opinion, my algorithm works the same as Alex Martelli’s, with all the benefits thereof, but has the added benefit highly efficient simple-test rejections that save a lot of time, not to mention the reduction in size of test numbers by powers of 4, which improves speed, efficiency, accuracy and the size of numbers that are testable. Probably especially true in non-Python implementations.

Roughly 99% of all integers are rejected as non-Square before Babylonian root extraction is even implemented, and in 2/3 the time it would take the Babylonian to reject the integer. And though these tests dont speed up the process that significantly, the reduction in all test numbers to an odd by dividing out all powers of 4 really accelerates the Babylonian test.

I did a time comparison test. I tested all integers from 1 to 10 Million in succession. Using just the Babylonian method by itself (with my specially tailored initial guess) it took my Surface 3 an average of 165 seconds (with 100% accuracy). Using just the logical tests in my algorithm (excluding the Babylonian), it took 127 seconds, it rejected 99% of all integers as non-Square without mistakenly rejecting any perfect squares. Of those integers that passed, only 3% were perfect Squares (a much higher density). Using the full algorithm above that employs both the logical tests and the Babylonian root extraction, we have 100% accuracy, and test completion in only 14 seconds. The first 100 Million integers takes roughly 2 minutes 45 seconds to test.

EDIT: I have been able to bring down the time further. I can now test the integers 0 to 100 Million in 1 minute 40 seconds. A lot of time is wasted checking the data type and the positivity. Eliminate the very first two checks and I cut the experiment down by a minute. One must assume the user is smart enough to know that negatives and floats are not perfect squares.

Answered By: CogitoErgoCogitoSum

I think that this works and is very simple:

import math

def is_square(num):
    sqrt = math.sqrt(num)
    return sqrt == int(sqrt)

It is incorrect for a large non-square such as 152415789666209426002111556165263283035677490.

Answered By: alejandro izquierdo

My answer is:

def is_square(x):
    return x**.5 % 1 == 0

It basically does a square root, then modulo by 1 to strip the integer part and if the result is 0 return True otherwise return False. In this case x can be any large number, just not as large as the max float number that python can handle: 1.7976931348623157e+308

It is incorrect for a large non-square such as 152415789666209426002111556165263283035677490.

Answered By: Lasagnenator

If the modulus (remainder) leftover from dividing by the square root is 0, then it is a perfect square.

def is_square(num: int) -> bool:
    return num % math.sqrt(num) == 0

I checked this against a list of perfect squares going up to 1000.

Answered By: GeneralCode

A variant of @Alex Martelli’s solution without set

When x in seen is True:

  • In most cases, it is the last one added, e.g. 1022 produces the x‘s sequence 511, 256, 129, 68, 41, 32, 31, 31;
  • In some cases (i.e., for the predecessors of perfect squares), it is the second-to-last one added, e.g. 1023 produces 511, 256, 129, 68, 41, 32, 31, 32.

Hence, it suffices to stop as soon as the current x is greater than or equal to the previous one:

def is_square(n):
    assert n > 1
    previous = n
    x = n // 2
    while x * x != n:
        x = (x + (n // x)) // 2
        if x >= previous:
            return False
        previous = x
    return True

x = 12345678987654321234567 ** 2
assert not is_square(x-1)
assert is_square(x)
assert not is_square(x+1)

Equivalence with the original algorithm tested for 1 < n < 10**7. On the same interval, this slightly simpler variant is about 1.4 times faster.

Answered By: Aristide
a=int(input('enter any number'))
for i in range(1,a):
    if a==i*i:
        print(a,'is perfect square number')
if flag==1:
    print(a,'is not perfect square number')
Answered By: Pemba Tamang

The idea is to run a loop from i = 1 to floor(sqrt(n)) then check if squaring it makes n.

bool isPerfectSquare(int n) 
    for (int i = 1; i * i <= n; i++) { 

        // If (i * i = n) 
        if ((n % i == 0) && (n / i == i)) { 
            return true; 
    return false; 
Answered By: Pragati Verma

It is possible to improve the Babylonian method by observing that the successive terms form a decreasing sequence if one starts above the square root of n.

def is_square(n):
    assert n > 1
    a = n
    b = (a + n // a) // 2
    while b < a:
        a = b
        b = (a + n // a) // 2
    return a * a == n
Answered By: Oblivier

If it’s a perfect square, its square root will be an integer, the fractional part will be 0, we can use modulus operator to check fractional part, and check if it’s 0, it does fail for some numbers, so, for safety, we will also check if it’s square of the square root even if the fractional part is 0.

import math

def isSquare(n):
    root = math.sqrt(n)
    if root % 1 == 0:
        if int(root) * int(root) == n:
            return True
    return False

Answered By: Abhishek Choudhary

In kotlin :

It’s quite easy and it passed all test cases as well.

really thanks to >>

fun isPerfectSquare(num: Int): Boolean {
    var result = false
    var sum=0L
    var oddNumber=1L
         sum = sum + oddNumber
         oddNumber = oddNumber+2
    result = sum == num.toLong()
 return result   
Answered By: Yogendra
def isPerfectSquare(self, num: int) -> bool:
    left, right = 0, num
    while left <= right:
        mid = (left + right) // 2
        if mid**2 < num:
            left = mid + 1
        elif mid**2 > num:
            right = mid - 1
            return True
    return False
Answered By: user12038667

A simple way to do it (faster than the second one) :

def is_square(n):  
     return str(n**(1/2)).split(".")[1] == '0'

Another way:

def is_square(n):   
    if n == 0:
        return True
        if n % 2 == 0 :
            for i in range(2,n,2):
                if i*i == n:
                    return True
        else :
            for i in range(1,n,2):
                if i*i == n:
                    return True
    return False
Answered By: bart-khalid

This is an elegant, simple, fast and arbitrary solution that works for Python version >= 3.8:

from math import isqrt

def is_square(number):
    if number >= 0:
        return isqrt(number) ** 2 == number
    return False
Answered By: Reza K Ghazi
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