Integer square root

In number theory, the integer square root of a non-negative integer is the non-negative integer which is the greatest integer less than or equal to the square root of,
For example,

Introductory remark

Let and be non-negative integers.
Algorithms that compute run forever on each input which is not a perfect square.
Algorithms that compute do not run forever. They are nevertheless capable of computing up to any desired accuracy.
Choose any and compute.
For example :
Compare the results with
It appears that the multiplication of the input by gives an accuracy of decimal digits.
To compute the decimal representation of, one can execute an infinite number of times, increasing by a factor at each pass.
Assume that in the next program the procedure is already defined and — for the sake of the argument — that all variables can hold integers of unlimited magnitude.
Then will print the entire decimal representation of.
import math # assume isqrt computation as given here
def sqrtForever:
""" Print sqrt, without halting """
result = math.isqrt
print # print result, followed by a decimal point
while True: # repeat forever...
y *= 100 # theoretical example: overflow is ignored
result = math.isqrt
print # print last digit of result

The conclusion is that algorithms which compute are computationally equivalent to algorithms which compute .
Another derivation of from is given in section Continued fraction of √c based on isqrt below.

Basic algorithms

The integer square root of a non-negative integer can be defined as
For example, because.

Algorithm using linear search

The following Python programs are straightforward implementations.

Linear search using addition

In the program above one can replace multiplication by addition, using the equivalence
def isqrt -> int:
"""
Integer square root
using addition
"""
L = 0
a = 1
d = 3
while a <= y:
a = a + d
d = d + 2
L = L + 1
return L

Algorithm using binary search

Linear search sequentially checks every value until it hits the smallest where.
A speed-up is achieved by using binary search instead.
def isqrt -> int:
""" Integer square root """
L = 0 # lower bound of the square root
R = y + 1 # upper bound of the square root
while :
M = // 2 # midpoint to test
if :
L = M
else:
R = M
return L

Numerical examples

,.
Using binary search the computation of converges to in iteration steps via the sequence
The computation of converges to in steps via the sequence

Algorithm using Newton's method

One way of calculating and is to use Heron's method, which is a special case of Newton's method, to find a solution for the equation, giving the iterative formula
The sequence converges quadratically to as.

Using only integer division

For computing one can use the quotient of Euclidean division for both of the division operations. This has the advantage of only using integers for each intermediate value, thus making the use of floating point representations unnecessary.
Let and initial guess. Define the integer sequence:

Proof of convergence

1. Positivity:
All terms are positive integers: for all.
2. Monotonicity:

If, then ;
If, then ;

3. Boundedness:
The sequence is bounded below by 1 and above by, so it is bounded.
4. Stabilization / Oscillation:
A bounded monotone integer sequence either stabilizes or oscillates between two consecutive integers:
5. Integer "Fixed-point" Condition:
At stabilization or oscillation:
6. Conclusion:
The sequence eventually stabilizes at or oscillates between and.
Remark:

is a strict fixed point unless is a perfect square.
If is a perfect square, the sequence oscillates between and.

Example implementation

def isqrt -> int:
"""
isqrt via Newton-Heron iteration with specified initial guess.
Uses 2-cycle oscillation detection.
Preconditions:
n >= 0 # isqrt = 0
x0 > 0, defaults to 1 # initial guess
Output:
isqrt
"""
assert n >= 0 and x0 > 0, "Invalid input"
# isqrt = 0; isqrt = 1
if n < 2: return n
prev2 = -1 # x_
prev1 = x0 # x_
while True:
x1 = // 2
# Case 1: converged
if x1 prev1:
return x1
# Case 2: oscillation
if x1 prev2 and x1 != prev1:
# We’re flipping between prev1 and prev2
# Choose the smaller one
return min
# Move forward
prev2, prev1 = prev1, x1

Numerical examples

The call isqrt converges to in 14 passes through while:
One iteration is gained by setting x0 to with the call isqrt. Although Heron's method converges quadratically close to the solution, less than one bit precision per iteration is gained at the beginning. This means that the choice of the initial estimate is critical for the performance of the algorithm. When a fast computation for the integer part of the binary logarithm or for the bit-length is available, one should better start at which is the least power of two bigger than. In the example of the integer square root of,,, and the resulting sequence is
In this case only four iteration steps are needed. This corresponds to the call isqrt.

Digit-by-digit algorithm

The traditional pen-and-paper algorithm for computing the square root is based on working from higher digit places to lower, and as each new digit pick the largest that will still yield a square. If stopping after the one's place, the result computed will be the integer square root.

Using bitwise operations

If working in base 2, the choice of digit is simplified to that between 0 and 1, and digit manipulations can be expressed in terms of binary shift operations. With * being multiplication, << being left shift, and >> being logical right shift, a recursive algorithm to find the integer square root of any natural number is:
def isqrt_recursive -> int:
assert n >= 0, "n must be a non-negative integer"
if n < 2: return n
# Recursive call:
small_cand = isqrt_recursive << 1 # same as 2 * isqrt
large_cand = small_cand + 1
if large_cand * large_cand > n: return small_cand
else: return large_cand

Equivalent non-recursive program:
def isqrt_iterative -> int:
""" Guy, Martin. "Fast integer square root by Mr. Woo's abacus algorithm" """
assert x >= 0, "x must be a non-negative integer"
op = x; res = 0
# note: i << 1 is 2 * i, i << 2 is 4 * i,
# i >> 1 is i // 2, and i >> 2 is i // 4...
# "one" starts at the highest power of four <= x
one = 1
while one <= op: one <<= 2
one >>= 2
while one != 0:
dltasqr = res + one
if op >= dltasqr:
op -= dltasqr
res += one << 1
res >>= 1
one >>= 2
return res

See for an example.

Karatsuba square root algorithm

The Karatsuba square root algorithm applies the same divide-and-conquer principle as the Karatsuba multiplication algorithm to compute integer square roots.
The method was formally analyzed by Paul Zimmermann. It recursively splits the input number into high and low halves, computes the square root of the higher half, and then determines the lower half algebraically.

Algorithm

Paul Zimmermann gives the following algorithm.
Because only one recursive call is made per level, the total complexity remains in the number of bits. Each level performs only linear-time arithmetic on half-size numbers.

Use and history

The Karatsuba-style square root is mainly used for arbitrary-precision arithmetic on very large integers, where it combines efficiently with and Karatsuba multiplication. It was first analyzed formally by Paul Zimmermann. Earlier practical work includes Martin Guy, and recursive versions appear in Donald Knuth. Modern GMP and MPIR libraries implement similar recursive techniques.

Implementation in Python

The Python program below implements Zimmermann’s algorithm. Given an integer, SqrtRem computes simultaneously its integer square root and the corresponding remainder. The choice of isqrt is ad libitum.
def SqrtRem -> tuple:
"""
Implementation based on Zimmermann's Karatsuba-style integer square root
algorithm . It recursively splits the input n into "limbs"
of size `word_bits` and combines partial results to compute the integer square root.
Args:
n : Non-negative integer to compute the square root of.
word_bits : Number of bits per "limb" or chunk
used when recursively splitting n. Default is 32. Each
limb represents a fixed-size part of n for the algorithm.
Returns:
tuple: s = integer square root of n, r = remainder.
Notes:
The limb size controls recursion granularity. Larger word_bits
reduces recursion depth but increases the size of subproblems;
smaller word_bits increases recursion depth but works on smaller chunks.
Reference:
Zimmermann, P.. "Karatsuba Square Root", Research report #3805, Inria.
Archived: https://inria.hal.science/inria-00072854v1/file/RR-3805.pdf
"""
if n < 0:
raise ValueError
if n 0:
return 0, 0 # trivial case
# Determine number of word-sized limbs
limblen = // word_bits
# Base case: single limb — compute directly
if limblen <= 1:
s = isqrt # any isqrt, e.g., math.isqrt or custom
r = n - s*s
return s, r
# --- Step 1: Split n into high and low parts ---
half_limbs = limblen // 2
shift = half_limbs * word_bits
hi = n >> shift # high half, corresponds to a3*b + a2
lo = n & # low half, corresponds to a1*b + a0
# --- Step 2: Recursive call on the high part ---
s_high, r_high = SqrtRem # approximate sqrt of high half
# --- Step 3: Recombine to approximate full sqrt ---
quarter = shift // 2
numerator = | # simulate Zimmermann’s DivRem step
denominator = s_high << 1 # 2*s' term
q = numerator // denominator if denominator else 0 # integer division
s_candidate = + q # recombine high and low
# --- Step 4: Verification and correction ---
# Ensure remainder is non-negative and s*s <= n < *
s = s_candidate
r = n - s*s
while r < 0: # overestimate correction
s -= 1
r = n - s*s
while * <= n: # underestimate correction
s += 1
r = n - s*s
return s, r

Example usage
for n in :
s, r = SqrtRem
print

Computation

SqrtRem = 65536, remainder = 5

SqrtRem = 3513641828, remainder = 5763386306

SqrtRem = 379032737378102767370356320425415662904513187772631008578870126471203845870697482014374611530431269030880793627229265919475483409207718357286202948008100864063587640630090308972232735749901964068667724412528434753635948938919935, remainder = 758065474756205534740712640850831325809026375545262017157740252942407691741394964028749223060862538061761587254458531838950966818415436714572405896016201728127175281260180617944465471499803928137335448825056869507271897877839870

In programming languages

Some programming languages dedicate an explicit operation to the integer square root calculation in addition to the general case or can be extended by libraries to this end.

Programming language	Example use	Version introduced
Chapel	`BigInteger.sqrt;` `BigInteger.sqrtRem;`	Unknown
Common Lisp		Unknown
Crystal	`Math.isqrt`	1.2
Java	`n.sqrt`	9
Julia	`isqrt`	0.3
Maple	`isqrt`	Unknown
PARI/GP	`sqrtint`	1.35a or before
PHP	`sqrt`	4
Python	`math.isqrt`	3.8
Racket		Unknown
Ruby	`Integer.sqrt`	2.5.0
Rust	`n.isqrt` `n.checked_isqrt`	1.84.0
SageMath	`isqrt`	Unknown
Scheme		R⁶RS
Tcl	`isqrt`	8.5
Zig	`std.math.sqrt`	Unknown

Continued fraction of √c based on isqrt

The computation of the simple continued fraction of can be carried out using only integer operations, with serving as the initial term. The algorithm generates continued fraction expansion in canonical form.
Let
be the integer square root of.
If is a perfect square, the continued fraction terminates immediately:
Otherwise, the continued fraction is periodic:
where the overline indicates the repeating part.
The continued fraction can be obtained by the following recurrence, which uses only integer arithmetic:
Since there are only finitely many possible triples, eventually one repeats, and from that point onward the continued fraction becomes periodic.

Implementation in Python

On input, a non-negative integer, the following program computes the simple continued fraction of. The integer square root is computed once. Only integer arithmetic is used. The program outputs, where the second element is the periodic part.
def continued_fraction_sqrt -> tuple where the second element is the periodic part.
For perfect squares, the period is empty.
"""
a0 = isqrt
# Perfect square: return period empty
if a0 * a0 c:
return )
m, d, a = 0, 1, a0
period =
seen = set
while True:
m_next = d * a - m
d_next = // d
a_next = // d_next
if in seen:
break
seen.add)
period.append
m, d, a = m_next, d_next, a_next
return

Example usage
for c in list + + :
cf = continued_fraction_sqrt
print

'''Output'''