Integer 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, isqrt ⁡ ( n ) = ⌊ n ⌋ . {\displaystyle \operatorname {isqrt} (n)=\lfloor {\sqrt {n}}\rfloor .}

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

Introductory remark

Let y {\displaystyle y} and k {\displaystyle k} be non-negative integers.

Algorithms that compute (the decimal representation of) y {\displaystyle {\sqrt {y}}} run forever on each input y {\displaystyle y} which is not a perfect square.

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

Choose any k {\displaystyle k} and compute ⌊ y × 100 k ⌋ {\textstyle \lfloor {\sqrt {y\times 100^{k}}}\rfloor }.

For example (setting y = 2 {\displaystyle y=2}): k = 0 : ⌊ 2 × 100 0 ⌋ = ⌊ 2 ⌋ = 1 k = 1 : ⌊ 2 × 100 1 ⌋ = ⌊ 200 ⌋ = 14 k = 2 : ⌊ 2 × 100 2 ⌋ = ⌊ 20000 ⌋ = 141 k = 3 : ⌊ 2 × 100 3 ⌋ = ⌊ 2000000 ⌋ = 1414 ⋮ k = 8 : ⌊ 2 × 100 8 ⌋ = ⌊ 20000000000000000 ⌋ = 141421356 ⋮ {\displaystyle {\begin{aligned}&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{aligned}}}

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

It appears that the multiplication of the input by 100 k {\displaystyle 100^{k}} gives an accuracy of k decimal digits.

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

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

Then sqrtForever ⁡ ( y ) {\displaystyle \operatorname {sqrtForever} (y)} will print the entire decimal representation of y {\displaystyle {\sqrt {y}}}.

importmath # assume isqrt computation as given here defsqrtForever(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 isqrt() are computationally equivalent to algorithms which compute sqrt().

Another derivation of y {\displaystyle {\sqrt {y}}} from ⌊ y ⌋ {\displaystyle \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 {\displaystyle y} can be defined as ⌊ y ⌋ = max { x : x 2 ≤ y < ( x + 1 ) 2 , x ∈ N } {\displaystyle \lfloor {\sqrt {y}}\rfloor =\max\{x:x^{2}\leq y<(x+1)^{2},x\in \mathbb {N} \}}

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

Algorithm using linear search

The following Python programs are straightforward implementations.

defisqrt(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 Ldefisqrt(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 + 2 L + 1 = L 2 + 1 + ∑ i = 1 L 2. {\displaystyle (L+1)^{2}=L^{2}+2L+1=L^{2}+1+\sum _{i=1}^{L}2.}

defisqrt(y: int) -> int: """ Integer square root (linear search, ascending) 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 x {\displaystyle x} where x 2 > y {\displaystyle x^{2}>y}.

A speed-up is achieved by using binary search instead.

defisqrt(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 <= y): L = M else: R = M return L

Numerical examples

• isqrt ⁡ ( 0 ) = 0 {\displaystyle \operatorname {isqrt} (0)=0}, isqrt ⁡ ( 1 ) = 1 {\displaystyle \operatorname {isqrt} (1)=1}.
• Using binary search the computation of isqrt ⁡ ( 131072 ) {\displaystyle \operatorname {isqrt} (131072)} converges to 362 {\displaystyle 362} in 17 {\displaystyle 17} iteration steps via the [ L , R ] {\displaystyle [L,R]} sequence

[ 0 , 131073 ] → [ 0 , 65536 ] → [ 0 , 32768 ] → [ 0 , 16384 ] → [ 0 , 8192 ] → [ 0 , 4096 ] → [ 0 , 2048 ] → [ 0 , 1024 ] → [ 0 , 512 ] → [ 256 , 512 ] → [ 256 , 384 ] → [ 320 , 384 ] → [ 352 , 384 ] → [ 352 , 368 ] → [ 360 , 368 ] → [ 360 , 364 ] → [ 362 , 364 ] → [ 362 , 363 ] {\displaystyle {\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 isqrt ⁡ ( 2000000 ) {\displaystyle \operatorname {isqrt} (2000000)} converges to 1414 {\displaystyle 1414} in 21 {\displaystyle 21} steps via the [ L , R ] {\displaystyle [L,R]} sequence

[ 0 , 2000001 ] → [ 0 , 1000000 ] → [ 0 , 500000 ] → [ 0 , 250000 ] → [ 0 , 125000 ] → [ 0 , 62500 ] → [ 0 , 31250 ] → [ 0 , 15625 ] → [ 0 , 7812 ] → [ 0 , 3906 ] → [ 0 , 1953 ] → [ 976 , 1953 ] → [ 976 , 1464 ] → [ 1220 , 1464 ] → [ 1342 , 1464 ] → [ 1403 , 1464 ] → [ 1403 , 1433 ] → [ 1403 , 1418 ] → [ 1410 , 1418 ] → [ 1414 , 1418 ] → [ 1414 , 1416 ] → [ 1414 , 1415 ] {\displaystyle {\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 {\displaystyle 0}) needs 1414 steps.

Algorithm using Newton's method

One way of calculating n {\displaystyle {\sqrt {n}}} and isqrt ⁡ ( n ) {\displaystyle \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 {\displaystyle x^{2}-n=0}, giving the iterative formula x k + 1 = 1 2 ( x k + n x k ) , k ≥ 0 , x 0 > 0. {\displaystyle x_{k+1}={\frac {1}{2}}\!\left(x_{k}+{\frac {n}{x_{k}}}\right),\quad k\geq 0,\quad x_{0}>0.}

The sequence { x k } {\displaystyle \{x_{k}\}} converges quadratically to n {\displaystyle {\sqrt {n}}} as k → ∞ {\displaystyle k\to \infty }.

Using only integer division

For computing ⌊ n ⌋ {\displaystyle \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 {\displaystyle n>0} and initial guess x 0 > 0 {\displaystyle x_{0}>0}. Define the integer sequence:

x k + 1 = ⌊ x k + ⌊ n / x k ⌋ 2 ⌋ , k = 0 , 1 , 2 , … {\displaystyle 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 {\displaystyle x_{k}>0} for all k {\displaystyle k}.

2. Monotonicity:

• If x k > n {\displaystyle x_{k}>{\sqrt {n}}}, then ⌊ n / x k ⌋ ≤ n / x k {\displaystyle \lfloor n/x_{k}\rfloor \leq n/x_{k}};

so x k + 1 = ⌊ x k + ⌊ n / x k ⌋ 2 ⌋ < x k + n / x k 2 < x k {\displaystyle x_{k+1}=\left\lfloor {\frac {x_{k}+\lfloor n/x_{k}\rfloor }{2}}\right\rfloor <{\frac {x_{k}+n/x_{k}}{2}}<x_{k}}.

Hence the sequence decreases.

• If x k < n {\displaystyle x_{k}<{\sqrt {n}}}, then ⌊ n / x k ⌋ ≥ n / x k − 1 {\displaystyle \lfloor n/x_{k}\rfloor \geq n/x_{k}-1};

so x k + 1 ≥ x k + n / x k − 1 2 > x k − 1 {\displaystyle x_{k+1}\geq {\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 {\displaystyle 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 {\displaystyle x_{k+1}=x_{k}} or x k + 1 = x k ± 1 {\displaystyle x_{k+1}=x_{k}\pm 1}.

5. Integer "Fixed-point" Condition: At stabilization or oscillation:

x k + 1 = ⌊ ( x k + ⌊ n / x k ⌋ ) / 2 ⌋ {\displaystyle x_{k+1}=\lfloor (x_{k}+\lfloor n/x_{k}\rfloor )/2\rfloor }.

This ensures that the sequence is either at ⌊ n ⌋ {\displaystyle \lfloor {\sqrt {n}}\rfloor } or oscillating between the two nearest integers around n {\displaystyle {\sqrt {n}}}.

6. Conclusion: The sequence eventually stabilizes at ⌊ n ⌋ {\displaystyle \lfloor {\sqrt {n}}\rfloor } or oscillates between ⌊ n ⌋ {\displaystyle \lfloor {\sqrt {n}}\rfloor } and ⌈ n ⌉ {\displaystyle \lceil {\sqrt {n}}\rceil }.

Remark:

• ⌊ n ⌋ {\displaystyle \lfloor {\sqrt {n}}\rfloor } is a strict fixed point unless n + 1 {\displaystyle n+1} is a perfect square.
• If n + 1 {\displaystyle n+1} is a perfect square, the sequence oscillates between ⌊ n ⌋ {\displaystyle \lfloor {\sqrt {n}}\rfloor } and ⌈ n ⌉ {\displaystyle \lceil {\sqrt {n}}\rceil }.

Example implementation

defisqrt(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" # isqrt(0) = 0; isqrt(1) = 1 if n < 2: return n prev2 = -1 # x_{i-2} prev1 = x0 # x_{i-1} while True: x1 = (prev1 + n // prev1) // 2 # Case 1: converged (steady value) if x1 == prev1: return x1 # Case 2: oscillation (2-cycle) if x1 == prev2 and x1 != prev1: # We’re flipping between prev1 and prev2 # Choose the smaller one (the true integer sqrt) return min(prev1, x1) # Move forward prev2, prev1 = prev1, x1

Numerical examples

The call isqrt(2000000) converges to 1414 {\displaystyle 1414} in 14 passes through while:

1000000 → 500001 → 250002 → 125004 → 62509 → 31270 → 15666 → 7896 → 4074 → 2282 → 1579 → 1422 → 1414 → 1414 {\displaystyle {\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 ⌊ n / 2 ⌋ {\displaystyle \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. 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 ⌊ ( log 2 ⁡ n ) / 2 ⌋ + 1 , {\displaystyle x_{0}=2^{\lfloor (\log _{2}n)/2\rfloor +1},} which is the least power of two bigger than n {\displaystyle {\sqrt {n}}}. In the example of the integer square root of 2000000, ⌊ log 2 ⁡ n ⌋ = 20 {\displaystyle \lfloor \log _{2}n\rfloor =20}, x 0 = 2 11 = 2048 {\displaystyle x_{0}=2^{11}=2048}, and the resulting sequence is 2048 → 1512 → 1417 → 1414 → 1414. {\displaystyle 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 n {\displaystyle {\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 ≤ n {\displaystyle \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 left shift, and >> being logical right shift, a recursive algorithm to find the integer square root of any natural number is:

defisqrt_recursive(n: int) -> int: assert n >= 0, "n must be a non-negative integer" if n < 2: return n # Recursive call: small_cand = isqrt_recursive(n >> 2) << 1 # same as 2 * isqrt(n // 4) large_cand = small_cand + 1 if large_cand * large_cand > n: return small_cand else: return large_cand

Equivalent non-recursive program:

defisqrt_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 # 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 Methods of computing square roots § Binary numeral system (base 2) 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.

Algorithm SqrtRem ( n = a 3 b 3 + a 2 b 2 + a 1 b + a 0 ) {\displaystyle {\text{Algorithm }}{\text{SqrtRem}}(n=a_{3}b^{3}+a_{2}b^{2}+a_{1}b+a_{0})}

Input: 0 ≤ a i < b , with a 3 ≥ b / 4 {\displaystyle {\text{Input: }}0\leq a_{i}<b,{\text{ with }}a_{3}\geq b/4}

Output: ( s , r ) such that s 2 ≤ n = s 2 + r < ( s + 1 ) 2 {\displaystyle {\text{Output: }}(s,r){\text{ such that }}s^{2}\leq n=s^{2}+r<(s+1)^{2}}

( s ′ , r ′ ) ← SqrtRem ( a 3 b + a 2 ) {\displaystyle \qquad (s',r')\gets {\text{SqrtRem}}(a_{3}b+a_{2})}

( q , u ) ← DivRem ( r ′ b + a 1 , 2 s ′ ) {\displaystyle \qquad (q,u)\gets {\text{DivRem}}(r'b+a_{1},2s')}

s ← s ′ b + q {\displaystyle \qquad s\gets s'b+q}

r ← u b + a 0 − q 2 {\displaystyle \qquad r\gets ub+a_{0}-q^{2}}

if r < 0 then {\displaystyle \qquad {\text{if }}r<0{\text{ then }}}

r ← r + 2 s − 1 {\displaystyle \qquad \qquad r\gets r+2s-1}

s ← s − 1 {\displaystyle \qquad \qquad s\gets s-1}

return ( s , r ) {\displaystyle \qquad {\text{return }}(s,r)}

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

Comparison with Karatsuba multiplication

Use and history

The Karatsuba-style square root is mainly used for arbitrary-precision arithmetic on very large integers, where it combines efficiently with Burnikel–Ziegler division[de] 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 ≥ 0 {\displaystyle n\geq 0}, SqrtRem computes simultaneously its integer square root s = ⌊ n ⌋ {\displaystyle s=\lfloor {\sqrt {n}}\rfloor } and the corresponding remainder r = n − s 2 {\displaystyle r=n-s^{2}}. The choice of isqrt() is ad libitum.

defSqrtRem(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 < 0: raise ValueError("n must be non-negative") if n == 0: return 0, 0 # trivial case # Determine number of word-sized limbs (mimics “limb” splitting in Zimmermann) limblen = (n.bit_length() + word_bits - 1) // word_bits # Base case: single limb — compute directly if limblen <= 1: s = isqrt(n) # 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 & ((1 << shift) - 1) # low half, corresponds to a1*b + a0 # --- Step 2: Recursive call on the high part --- s_high, r_high = SqrtRem(hi, word_bits) # approximate sqrt of high half # --- Step 3: Recombine to approximate full sqrt --- quarter = shift // 2 numerator = (r_high << quarter) | (lo >> quarter) # simulate Zimmermann’s DivRem step denominator = s_high << 1 # 2*s' term q = numerator // denominator if denominator else 0 # integer division s_candidate = (s_high << quarter) + q # recombine high and low # --- Step 4: Verification and correction --- # Ensure remainder is non-negative and s*s <= n < (s+1)*(s+1) s = s_candidate r = n - s*s while r < 0: # overestimate correction s -= 1 r = n - s*s while (s + 1)*(s + 1) <= n: # underestimate correction s += 1 r = n - s*s return s, r

Example usage

for n in [(2**32) + 5, 12345678901234567890, (1 << 1512) - 1]: s, r = SqrtRem(n) print(f"SqrtRem({n}) ={s}, remainder ={r}")

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.

Continued fraction of √c based on isqrt

The computation of the simple continued fraction of c {\displaystyle {\sqrt {c}}} can be carried out using only integer operations, with isqrt ⁡ ( c ) {\displaystyle \operatorname {isqrt} (c)} serving as the initial term. The algorithm generates continued fraction expansion in canonical form.

Let a 0 = ⌊ c ⌋ {\displaystyle a_{0}=\lfloor {\sqrt {c}}\rfloor } be the integer square root of c {\displaystyle c}.

If c {\displaystyle c} is a perfect square, the continued fraction terminates immediately:

c = [ a 0 ] . {\displaystyle {\sqrt {c}}=[a_{0}].}

Otherwise, the continued fraction is periodic:

c = [ a 0 ; a 1 , a 2 , … , a m ¯ ] {\displaystyle {\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 , d 0 = 1 , a 0 = ⌊ c ⌋ . {\displaystyle m_{0}=0,\quad d_{0}=1,\quad a_{0}=\lfloor {\sqrt {c}}\rfloor .}

For k ≥ 0 {\displaystyle k\geq 0},

m k + 1 = d k a k − m k , d k + 1 = c − m k + 1 2 d k , a k + 1 = ⌊ a 0 + m k + 1 d k + 1 ⌋ . {\displaystyle 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 ) {\displaystyle (m_{k},d_{k},a_{k})}, eventually one repeats, and from that point onward the continued fraction becomes periodic.

Implementation in Python

On input c {\displaystyle c}, a non-negative integer, the following program computes the simple continued fraction of c {\displaystyle {\sqrt {c}}}. The integer square root ⌊ c ⌋ {\displaystyle \lfloor {\sqrt {c}}\rfloor } is computed once. Only integer arithmetic is used. The program outputs [ a 0 , ( a 1 , a 2 , . . . , a m ) ] {\displaystyle [a0,(a1,a2,...,am)]}, where the second element is the periodic part.

defcontinued_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) # 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

Input (c) Output (cf) Continued fraction 0 [0] 0 = 0 {\displaystyle {\sqrt {0}}=0} 1 [1] 1 = 1 {\displaystyle {\sqrt {1}}=1} 2 [1; (2,)] 2 = [ 1 ; 2 ¯ ] {\displaystyle {\sqrt {2}}=[1;{\overline {2}}]} 3 [1; (1, 2)] 3 = [ 1 ; 1 , 2 ¯ ] {\displaystyle {\sqrt {3}}=[1;{\overline {1,2}}]} 4 [2] 4 = 2 {\displaystyle {\sqrt {4}}=2} 5 [2; (4,)] 5 = [ 2 ; 4 ¯ ] {\displaystyle {\sqrt {5}}=[2;{\overline {4}}]} 6 [2; (2, 4)] 6 = [ 2 ; 2 , 4 ¯ ] {\displaystyle {\sqrt {6}}=[2;{\overline {2,4}}]} 7 [2; (1, 1, 1, 4)] 7 = [ 2 ; 1 , 1 , 1 , 4 ¯ ] {\displaystyle {\sqrt {7}}=[2;{\overline {1,1,1,4}}]} 8 [2; (1, 4)] 8 = [ 2 ; 1 , 4 ¯ ] {\displaystyle {\sqrt {8}}=[2;{\overline {1,4}}]} 9 [3] 9 = 3 {\displaystyle {\sqrt {9}}=3} 10 [3; (6,)] 10 = [ 3 ; 6 ¯ ] {\displaystyle {\sqrt {10}}=[3;{\overline {6}}]} 11 [3; (3, 6)] 11 = [ 3 ; 3 , 6 ¯ ] {\displaystyle {\sqrt {11}}=[3;{\overline {3,6}}]} 12 [3; (2, 6)] 12 = [ 3 ; 2 , 6 ¯ ] {\displaystyle {\sqrt {12}}=[3;{\overline {2,6}}]} 13 [3; (1, 1, 1, 1, 6)] 13 = [ 3 ; 1 , 1 , 1 , 1 , 6 ¯ ] {\displaystyle {\sqrt {13}}=[3;{\overline {1,1,1,1,6}}]} 14 [3; (1, 2, 1, 6)] 14 = [ 3 ; 1 , 2 , 1 , 6 ¯ ] {\displaystyle {\sqrt {14}}=[3;{\overline {1,2,1,6}}]} 15 [3; (1, 6)] 15 = [ 3 ; 1 , 6 ¯ ] {\displaystyle {\sqrt {15}}=[3;{\overline {1,6}}]} 16 [4] 16 = 4 {\displaystyle {\sqrt {16}}=4} 17 [4; (8,)] 17 = [ 4 ; 8 ¯ ] {\displaystyle {\sqrt {17}}=[4;{\overline {8}}]} 114 [10; (1, 2, 10, 2, 1, 20)] 114 = [ 10 ; 1 , 2 , 10 , 2 , 1 , 20 ¯ ] {\displaystyle {\sqrt {114}}=[10;{\overline {1,2,10,2,1,20}}]} 4097280036 [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 terms

See also

• Square root algorithms
• Periodic continued fraction

Notes

External links

• Jarvis, Ashley Frazer (2006). (PDF). Mathematical Spectrum. 37: 119–122.
• Minsky, Marvin (1967). "9. The Computable Real Numbers". . Prentice-Hall. ISBN 0-13-165563-9. OCLC .
• .