Integer square root

---

title: "Integer square root" type: doc version: 1 created: 2026-02-28 author: "Wikipedia contributors" status: active scope: public tags: ["number-theoretic-algorithms", "number-theory", "root-finding-algorithms", "articles-with-example-c-code", "articles-with-example-python-(programming-language)-code", "articles-with-example-rust-code"] description: "Greatest integer less than or equal to square root" topic_path: "science/mathematics" source: "https://en.wikipedia.org/wiki/Integer_square_root" license: "CC BY-SA 4.0" wikipedia_page_id: 0 wikipedia_revision_id: 0

::summary Greatest integer less than or equal to square root ::

In number theory, the integer square root (isqrt) of a non-negative integer n is the non-negative integer m which is the greatest integer less than or equal to the square root of n, \operatorname{isqrt}(n) = \lfloor \sqrt n \rfloor.

For example, \operatorname{isqrt}(27) = \lfloor \sqrt{27} \rfloor = \lfloor 5.19615242270663 ... \rfloor = 5.

Introductory remark

Let y and k be non-negative integers.

Algorithms that compute (the decimal representation of) \sqrt y run forever on each input y which is not a perfect square.The square roots of the perfect squares (e.g., 0, 1, 4, 9, 16) are integers. In all other cases, the square roots of positive integers are irrational numbers.

Algorithms that compute \lfloor \sqrt y \rfloor do not run forever. They are nevertheless capable of computing \sqrt y up to any desired accuracy k.

Choose any k and compute \lfloor \sqrt {y \times 100^k} \rfloor.

For example (setting y = 2): \begin{align} & k = 0: \lfloor \sqrt {2 \times 100^{0}} \rfloor = \lfloor \sqrt {2} \rfloor = 1 \ & k = 1: \lfloor \sqrt {2 \times 100^{1}} \rfloor = \lfloor \sqrt {200} \rfloor = 14 \ & k = 2: \lfloor \sqrt {2 \times 100^{2}} \rfloor = \lfloor \sqrt {20000} \rfloor = 141 \ & k = 3: \lfloor \sqrt {2 \times 100^{3}} \rfloor = \lfloor \sqrt {2000000} \rfloor = 1414 \ & \vdots \ & k = 8: \lfloor \sqrt {2 \times 100^{8}} \rfloor = \lfloor \sqrt {20000000000000000} \rfloor = 141421356 \ & \vdots \ \end{align}

Compare the results with \sqrt {2} = 1.41421356237309504880168872420969807856967187537694 ...

It appears that the multiplication of the input by 100^k gives an accuracy of k decimal digits.It is no surprise that the repeated multiplication by 100 is a feature in

To compute the (entire) decimal representation of \sqrt y, one can execute \operatorname{isqrt}(y) an infinite number of times, increasing y by a factor 100 at each pass.

Assume that in the next program (\operatorname{sqrtForever}) the procedure \operatorname{isqrt}(y) is already defined and — for the sake of the argument — that all variables can hold integers of unlimited magnitude.

Then \operatorname{sqrtForever}(y) will print the entire decimal representation of \sqrt y.The fractional part of square roots of perfect squares is rendered as 000....

: import math # assume isqrt computation as given here

def sqrtForever(y: int): """ Print sqrt(y), without halting """ result = math.isqrt(y) print(str(result) + ".", end="") # print result, followed by a decimal point

while True: # repeat forever ... y *= 100 # theoretical example: overflow is ignored result = math.isqrt(y) print(str(result % 10), end="") # print last digit of result

The conclusion is that algorithms which compute are computationally equivalent to algorithms which compute .

Another derivation of \sqrt y from \lfloor \sqrt y \rfloor is given in section Continued fraction of √c based on isqrt below.

Basic algorithms

The integer square root of a non-negative integer y can be defined as \lfloor \sqrt y \rfloor = \max { x : x^2 \leq y

For example, \operatorname{isqrt}(27) = \lfloor \sqrt{27} \rfloor = 5 because 6^2 27 \text{ and } 5^2 \ngtr 27.

Algorithm using linear search

The following Python programs are straightforward implementations.

:{{flex columns |1= ::code[lang=python] def isqrt(y: int) -> int: """ Integer square root (linear search, ascending) """ # initial underestimate, L <= isqrt(y)
L = 0 while (L + 1) * (L + 1) <= y: L += 1

return L

::

|2= ::code[lang=python] def isqrt(y: int) -> int: """ Integer square root (linear search, descending) """ # initial overestimate, isqrt(y) <= R
R = y while (R * R > y): R -= 1

return R

::

Linear search using addition

In the program above (linear search, ascending) one can replace multiplication by addition, using the equivalence (L+1)^2 = L^2 + 2L + 1 = L^2 + 1 + \sum_{i=1}^L 2. : def isqrt(y: int) - int: """ Integer square root (linear search, ascending) using addition """ L = 0 a = 1 d = 3

while a 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 x where x^2 y.

A speed-up is achieved by using binary search instead. : def isqrt(y: int) - int: """ Integer square root (binary search) """ L = 0 # lower bound of the square root R = y + 1 # upper bound of the square root

while (L != R - 1): M = (L + R) // 2 # midpoint to test if (M * M L = M else: R = M

return L

Numerical examples

• \operatorname{isqrt}(0) = 0, \operatorname{isqrt}(1) = 1.
• Using binary search the computation of \operatorname{isqrt}(131072) converges to 362 in 17 iteration steps via the [L,R] sequence :\begin{aligned} &[0,131073] \to [0,65536] \to [0,32768] \to [0,16384] \to [0,8192] \to [0,4096] \rightarrow [0,2048] \to [0,1024] \to [0,512] \& \to [256,512] \to [256,384] \to [320,384] \to [352,384] \to [352,368] \to [360,368] \to [360,364] \to [362,364] \to [362,363] \end{aligned}
• The computation of \operatorname{isqrt}(2000000) converges to 1414 in 21 steps via the [L,R] sequence :\begin{aligned} & [0,2000001] \to [0,1000000] \to [0,500000] \to [0,250000] \to [0,125000] \to [0,62500] \to [0,31250] \to [0,15625] \& \to [0,7812] \to [0,3906] \to [0,1953] \to [976,1953] \to [976,1464] \to [1220,1464] \to [1342,1464] \to [1403,1464] \& \to [1403,1433] \to [1403,1418] \to [1410,1418] \to [1414,1418] \to [1414,1416] \to [1414,1415] \end{aligned} : Linear search (ascending, starting from 0) needs steps.

Algorithm using Newton's method

One way of calculating \sqrt{n} and \operatorname{isqrt}(n) is to use Heron's method, which is a special case of Newton's method, to find a solution for the equation x^2 - n = 0, giving the iterative formula x_{k+1} = \frac{1}{2}!\left(x_k + \frac{n}{x_k}\right), \quad k \ge 0, \quad x_0 0.

The sequence {x_k} converges quadratically to \sqrt{n} as k\to\infty.See Heron's method.

Using only integer division

For computing \lfloor \sqrt n \rfloor 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 n 0 and initial guess x_0 0. Define the integer sequence: :x_{k+1} = \left\lfloor \frac{x_k + \left\lfloor n / x_k \right\rfloor}{2} \right\rfloor, \quad k = 0, 1, 2, \dots

Proof of convergence

1. Positivity: All terms are positive integers: x_k 0 for all k.

2. Monotonicity:

• If x_k \sqrt{n}, then \lfloor n / x_k \rfloor \le n / x_k; : so x_{k+1} = \left\lfloor \frac{x_k + \lfloor n / x_k \rfloor}{2} \right\rfloor . :Hence the sequence decreases.

• If x_k , then \lfloor n / x_k \rfloor \ge n / x_k - 1; :so x_{k+1} \ge \frac{x_k + n / x_k - 1}{2} x_k - 1. :Hence the sequence increases or stays the same.

3. Boundedness: The sequence is bounded below by 1 and above by x_0, so it is bounded.

4. Stabilization / Oscillation: A bounded monotone integer sequence either stabilizes or oscillates between two consecutive integers: : x_{k+1} = x_k or x_{k+1} = x_k \pm 1.

5. Integer "Fixed-point" Condition: At stabilization or oscillation: : x_{k+1} = \lfloor (x_k + \lfloor n / x_k \rfloor)/2 \rfloor. :This ensures that the sequence is either at \lfloor \sqrt n \rfloor or oscillating between the two nearest integers around \sqrt n.

6. Conclusion: The sequence eventually stabilizes at \lfloor \sqrt n \rfloor or oscillates between \lfloor \sqrt n \rfloor and \lceil \sqrt n \rceil.

Remark:

• \lfloor \sqrt n \rfloor is a strict fixed point unless n + 1 is a perfect square.
• If n + 1 is a perfect square, the sequence oscillates between \lfloor \sqrt n \rfloor and \lceil \sqrt n \rceil.

Example implementation

: def isqrt(n: int, x0: int = 1) - int: """ isqrt via Newton-Heron iteration with specified initial guess. Uses 2-cycle oscillation detection.

Preconditions: n = 0 # isqrt(0) = 0 x0 0, defaults to 1 # initial guess

Output: isqrt(n) """ assert n = 0 and x0 0, "Invalid input"

1. isqrt(0) = 0; isqrt(1) = 1 if n

prev2 = -1 # x_{i-2} prev1 = x0 # x_{i-1}

while True: x1 = (prev1 + n // prev1) // 2

1. Case 1: converged (steady value) if x1 == prev1: return x1

2. Case 2: oscillation (2-cycle) if x1 == prev2 and x1 != prev1:

3. We’re flipping between prev1 and prev2

4. Choose the smaller one (the true integer sqrt) return min(prev1, x1)

5. Move forward prev2, prev1 = prev1, x1

Numerical examples

The call isqrt(2000000) converges to 1414 in 14 passes through while: :\begin{aligned} & 1000000 \to 500001 \to 250002 \to 125004 \to 62509 \to 31270 \to 15666 \to 7896 \to 4074 \to 2282 \& \to 1579 \to 1422 \to 1414 \rightarrow 1414\end{aligned}. One iteration is gained by setting x0 to \lfloor n/2 \rfloor with the call isqrt(2000000, 1000000). 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.Newton's method can be given as follows (with the initial guess set to s): ::code[lang=python] def isqrt(s: int) -> int: """isqrt via Newton/Heron iteration.""" L, R = 1, s while L < R: R = L + ((R - L) // 2) L = s // R return R ::

Computations

• \operatorname{isqrt}(0) = 0, \operatorname{isqrt}(1) = 1.
• The computation of \operatorname{isqrt}(2000000) consists in 13 ; [L,R] steps: :\begin{aligned} &[1,2000000] \to [2,1000000] \to [3,500001] \& \to [7,250002] \to [15,125004] \to [31,62509] \& \to [63,31270] \to [127,15666] \to [253,7896] \& \to [490,4074] \to [876,2282] \to [1266,1579] \& \to [1406,1422] \to [1414,1414] \end{aligned} :One sees that the performance gain of Newton's method over binary search is due to the fact that \lfloor \sqrt s \rfloor is approached simultaneously from Left and Right, whereas binary search adjusts only one side at each iteration. When a fast computation for the integer part of the binary logarithm or for the bit-length is available (like e.g. n.bit_length() in Python), one should better start at x_0 = 2^{\lfloor (\log_2 n) / 2 \rfloor + 1}, which is the least power of two bigger than \sqrt n. In the example of the integer square root of 2000000, \lfloor \log_2 n \rfloor = 20, x_0 = 2^{11} = 2048, and the resulting sequence is 2048 \rightarrow 1512 \rightarrow 1417 \rightarrow 1414 \rightarrow 1414. In this case only four iteration steps are needed. This corresponds to the call isqrt(2000000, 2048).

Digit-by-digit algorithm

The traditional pen-and-paper algorithm for computing the square root \sqrt{n} is based on working from higher digit places to lower, and as each new digit pick the largest that will still yield a square \leq n. 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 (the "small candidate") and 1 (the "large candidate"), and digit manipulations can be expressed in terms of binary shift operations. With * being multiplication, `` being logical right shift, a recursive algorithm to find the integer square root of any natural number is: : def isqrt_recursive(n: int) - int: assert n = 0, "n must be a non-negative integer" if n

1. Recursive call: small_cand = isqrt_recursive(n 2) large_cand = small_cand + 1 if large_cand * large_cand n: return small_cand else: return large_cand Equivalent non-recursive program:The algorithm is explained in Square_root_algorithms#Binary numeral system (base 2) : def isqrt_iterative(x: int) - int: """ Guy, Martin (1985). "Fast integer square root by Mr. Woo's abacus algorithm" """ assert x = 0, "x must be a non-negative integer"

op = x; res = 0

1. note: i

2. i 1 is i // 2, and i 2 is i // 4...

3. "one" starts at the highest power of four one = 1 while one one = 2

while one != 0: dltasqr = res + one if op = dltasqr: op -= dltasqr res += one 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 (1999). 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 (1999) gives the following algorithm. :\text{Algorithm } \text{SqrtRem}(n = a_3 b^3 + a_2 b^2 + a_1 b + a_0) :\text{Input: } 0 \le a_i :\text{Output: } (s,r) \text{ such that } s^2 \le n = s^2 + r :\qquad (s', r') \gets \text{SqrtRem}(a_3 b + a_2) :\qquad (q, u) \gets \text{DivRem}(r' b + a_1, 2 s') :\qquad s \gets s' b + q :\qquad r \gets u b + a_0 - q^2 :\qquad \text{if } r :\qquad \qquad r \gets r + 2s - 1 :\qquad \qquad s \gets s - 1 :\qquad \text{return } (s,r)

Because only one recursive call is made per level, the total complexity remains O(n) in the number of bits. Each level performs only linear-time arithmetic on half-size numbers.

Comparison with Karatsuba multiplication

::data[format=table]

Property	Karatsuba multiplication	Karatsuba-style square root
Recursive calls per level	3	1
Recurrence	T(n)=3T(n/2)+O(n)	T(n)=T(n/2)+O(n)
Asymptotic complexity	O(n^{\log_2 3})!\approx!O(n^{1.585})	O(n)
Key operation	Three partial multiplications and recombination	One recursive square root and algebraic correction
::

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 (1999). Earlier practical work includes Martin Guy (1985), and recursive versions appear in Donald Knuth (1998). Modern GMP and MPIR libraries implement similar recursive techniques.

Implementation in Python

The Python program below implements Zimmermann’s algorithm. Given an integer n \ge 0, SqrtRem computes simultaneously its integer square root s = \lfloor \sqrt{n} \rfloor and the corresponding remainder r = n - s^2. The choice of isqrt() is ad libitum. : def SqrtRem(n: int, word_bits: int = 32) - tuple[int, int]: """ Implementation based on Zimmermann's Karatsuba-style integer square root algorithm [Zimmermann, 1999]. It recursively splits the input n into "limbs" of size word_bits and combines partial results to compute the integer square root.

Args: n (int): Non-negative integer to compute the square root of. word_bits (int, optional): 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[int, int]: s = integer square root of n, r = remainder (n - s*s).

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. (1999). "Karatsuba Square Root", Research report #3805, Inria. Archived: https://inria.hal.science/inria-00072854v1/file/RR-3805.pdf """ if n raise ValueError("n must be non-negative") if n == 0: return 0, 0 # trivial case

1. Determine number of word-sized limbs (mimics “limb” splitting in Zimmermann) limblen = (n.bit_length() + word_bits - 1) // word_bits

2. Base case: single limb — compute directly if limblen s = isqrt(n) # any isqrt, e.g., math.isqrt or custom r = n - s*s return s, r

3. --- 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 & ((1

4. --- Step 2: Recursive call on the high part --- s_high, r_high = SqrtRem(hi, word_bits) # approximate sqrt of high half

5. --- Step 3: Recombine to approximate full sqrt --- quarter = shift // 2 numerator = (r_high quarter) # simulate Zimmermann’s DivRem step denominator = s_high

q = numerator // denominator if denominator else 0 # integer division s_candidate = (s_high

1. --- Step 4: Verification and correction ---
2. Ensure remainder is non-negative and ss s = s_candidate r = n - ss

while r s -= 1 r = n - ss while (s + 1)(s + 1) s += 1 r = n - s*s

return s, r

Example usage : for n in [(2**32) + 5, 12345678901234567890, (1 s, r = SqrtRem(n) print(f"SqrtRem({n}) = {s}, remainder = {r}") ::data[format=table]

Computation
::

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.

::data[format=table]

Programming language	Example use	Version introduced
Chapel	BigInteger.sqrt(result, n);
BigInteger.sqrtRem(result, remainder, n);	Unknown
Common Lisp	(isqrt n)	Unknown
Crystal	Math.isqrt(n)	1.2
Java	n.sqrt() (BigInteger only)	9
Julia	isqrt(n)	0.3
Maple	isqrt(n)	Unknown
PARI/GP	sqrtint(n)	1.35a (as isqrt) or before
PHP	sqrt($num)	4
Python	math.isqrt(n)	3.8
Racket	(integer-sqrt n)
(integer-sqrt/remainder n)	Unknown
Ruby	Integer.sqrt(n)	2.5.0
Rust	n.isqrt()
n.checked_isqrt()	1.84.0
SageMath	isqrt(n)	Unknown
Scheme	(exact-integer-sqrt n)	R6RS
Tcl	isqrt($n)	8.5
Zig	std.math.sqrt(n)	Unknown
::

Continued fraction of √c based on isqrt

The computation of the simple continued fraction of \sqrt{c} can be carried out using only integer operations, with \operatorname{isqrt}(c) serving as the initial term. The algorithm{{cite web |last= Beceanu |first= Marius |date= 2003-02-05 |title= Period of the Continued Fraction of sqrt(n) |url= https://web.math.princeton.edu/mathlab/jr02fall/Periodicity/mariusjp.pdf |at= Theorem 2.3 |access-date= 2025-10-05 |archive-url= https://web.archive.org/web/20151221205104/http://web.math.princeton.edu/mathlab/jr02fall/Periodicity/mariusjp.pdf |url-status=live |archive-date=2015-12-21

Let a_0 = \lfloor \sqrt{c} \rfloor be the integer square root of c.

If c is a perfect square, the continued fraction terminates immediately: :\sqrt{c} = [a_0]. Otherwise, the continued fraction is periodic: :\sqrt{c} = [a_0; \overline{a_1, a_2, \dots, a_m}], where the overline indicates the repeating part.

The continued fraction can be obtained by the following recurrence, which uses only integer arithmetic: :m_{0} = 0, \quad d_{0} = 1, \quad a_{0} = \lfloor \sqrt{c} \rfloor. :For k \geq 0, : m_{k+1} = d_{k} a_{k} - m_{k}, \quad d_{k+1} = \frac{c - m_{k+1}^{2}}{d_{k}}, \quad a_{k+1} = \left\lfloor \frac{a_{0} + m_{k+1}}{d_{k+1}} \right\rfloor.

Since there are only finitely many possible triples (m_{k}, d_{k}, a_{k}), eventually one repeats, and from that point onward the continued fraction becomes periodic.

Implementation in Python

continued_fraction_sqrt_Python On input c, a non-negative integer, the following program computes the simple continued fraction of \sqrt{c}. The integer square root \lfloor \sqrt{c} \rfloor is computed once. Only integer arithmetic is used. The program outputs [a0, (a1, a2, ..., am)], where the second element is the periodic part. : def continued_fraction_sqrt(c: int) - tuple[int, tuple[int, ...]]: """ Compute the continued fraction of sqrt(c) using integer arithmetic. Returns [a0, (a1, a2, ..., am)] where the second element is the periodic part. For perfect squares, the period is empty. """ a0 = isqrt(c)

1. Perfect square: return period empty if a0 * a0 == c: return (a0, ())

m, d, a = 0, 1, a0 period = [] seen = set()

while True: m_next = d * a - m d_next = (c - m_next * m_next) // d a_next = (a0 + m_next) // d_next

if (m_next, d_next, a_next) in seen: break

seen.add((m_next, d_next, a_next)) period.append(a_next) m, d, a = m_next, d_next, a_next

return (a0, tuple(period))

Example usage : for c in list(range(0, 18)) + [114] + [4097280036]: cf = continued_fraction_sqrt(c) print(f"sqrt({c}): {cf}") Output :{| class="wikitable" ! Input (c) ! Output (cf) ! Continued fraction |- | 0 || [0] || \sqrt{0} = 0 |- | 1 || [1] || \sqrt{1} = 1 |- | 2|| [1; (2,)] || \sqrt{2} = [1; \overline{2}] |- | 3|| [1; (1, 2)] || \sqrt{3} = [1; \overline{1,2}] |- | 4 || [2] || \sqrt{4} = 2 |- | 5 || [2; (4,)] || \sqrt{5} = [2; \overline{4}] |- | 6 || [2; (2, 4)] || \sqrt{6} = [2; \overline{2,4}] |- | 7 || [2; (1, 1, 1, 4)] || \sqrt{7} = [2; \overline{1,1,1,4}] |- | 8 || [2; (1, 4)] || \sqrt{8} = [2; \overline{1,4}] |- | 9 || [3] || \sqrt{9} = 3 |- | 10 || [3; (6,)] || \sqrt{10} = [3; \overline{6}] |- | 11 || [3; (3, 6)] || \sqrt{11} = [3; \overline{3,6}] |- | 12 || [3; (2, 6)] || \sqrt{12} = [3; \overline{2,6}] |- | 13 || [3; (1, 1, 1, 1, 6)] || \sqrt{13} = [3; \overline{1, 1, 1, 1, 6}] |- | 14 || [3; (1, 2, 1, 6)] || \sqrt{14} = [3; \overline{1, 2, 1, 6}] |- | 15 || [3; (1, 6)] || \sqrt{15} = [3; \overline{1,6}] |- | 16 || [4] || \sqrt{16} = 4 |- | 17 || [4; (8,)] || \sqrt{17} = [4; \overline{8}] |- | 114 || [10; (1, 2, 10, 2, 1, 20)] || \sqrt{114} = [10; \overline{1,2,10,2,1,20}]See the example in the article Periodic continued fraction. |- | 4097280036 || colspan="2" | [64009; (1, 1999, 3, 4, 1, 499, 3, 1, 3, 3, 1, 124, ... ..., 3, 1, 3, 499, 1, 4, 3, 1999, 1, 128018)] period: 13,032 termsThe continued fraction expansion of \sqrt{4097280036} has a period of 13,032 terms. While Python is unable to display the entire sequence on screen due to its length, writing the output to a file completes successfully. |}

Notes

References

References

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2. Guy, Martin. (1985). "Fast integer square root by Mr. Woo's abacus algorithm". University of Kent at Canterbury (UKC).
3. Zimmermann, Paul. (1999 ). "Karatsuba Square Root". [[French Institute for Research in Computer Science and Automation.
4. Knuth, Donald E.. (1998). "The Art of Computer Programming, Volume 2: Seminumerical Algorithms". Addison–Wesley.
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