Reference · Math

math.isqrt

math.isqrt(int) -> int

What it does

The integer square root: the largest int whose square is not above n, as Python's math.isqrt. math.isqrt(17) is 4. It is exact for every int, where math.floor(math.sqrt(n)) can be off by one for ints past 2⁵³.

Examples

print(math.isqrt(0), math.isqrt(1), math.isqrt(15), math.isqrt(16), math.isqrt(17))print(type_of(math.isqrt(17)), math.sqrt(17))

0 1 3 4 4int 4.123105625617661

Edge cases

• Exact up to the largest int: math.isqrt(9223372036854775807) is 3037000499.
• A perfect square gives its root exactly; math.isqrt(n) * math.isqrt(n) == n tests for one.
• A float that is a whole number is accepted: math.isqrt(16.0) is 4.

print(math.isqrt(9223372036854775807))let square = (n) => math.isqrt(n) * math.isqrt(n) == nprint(square(144), square(145), math.isqrt(16.0))

3037000499true false 4

Errors

A negative number is a runtime error, not an error value, with a message that says so, as Python's ValueError. A fraction, text or null is a runtime error as well.

print(math.isqrt(-1))

line 1: `math.isqrt` is not defined for negative numbers (-1)

print(math.isqrt(2.5))

line 1: `math.isqrt` expects an int as argument 1, got float

Chaining

In a pipeline the number comes first:

print(1000000 | math.isqrt)print([1, 2, 3, 4, 5] | map(math.isqrt))

1000[1, 1, 1, 2, 2]

See also

math.sqrt, math.floor, math.gcd

Edit this page on GitHub · every example's output was made by running it (tools/refcheck.mjs) · questions on Discord or r/quillo