Half code required for this question for complete. Given an integer num_perfect_squares will return the minimum amount of perfect squares are requiredto sum to the specified number. Lagrange's four-square theorem gives us that the answer will alwaysbe between 1 and 4 (https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem).Some examples:Number | Perfect Squares representation | Answer-|---|------9|3^210| 3^2+1^212 12^2 + 2^2 + 2^231 | 5^2 + 2^2+1^2+1^2| 1return 1121314import mathdef num_perfect_squares(number):Returns the smallest number of perfect squares that sum to the specified number.:return: int between 1 - 4# If the number is a perfect square then we only need 1 number.if int(math.sqrt(number))**2 :==number:#We check if https://en.wikipedia.org/wiki/Legendre%27s_three-square_theorem holds and divide# the number accordingly. le. if the number can be written as a sum of 3 squares (where the# 0^2 is allowed), which is possible for all numbers except those of the form: 4^a(8b + 7).while number > 0 and number % 4 == 0:number /= 4# If the number is of the form: 4^a(8b + 7) it can't be expressed as a sum of three (or less# excluding the 0^2) perfect squares. If the number was of that form, the previous while loop# divided away the 4^a, so by now it would be of the form: 8b + 7. So check if this is the case# and return 4 since it neccessarily must be a sum of 4 perfect squares, in accordance# with https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem.if number % 8 == 7:return 4# By now we know that the number wasn't of the form 4^a(8b + 7) so it can be expressed as a sum# of 3 or less perfect squares. Try first to express it as a sum of 2 perfect squares, and if# that fails, we know finally that it can be expressed as a sum of 3 perfect squares.for i in range(1, int(math.sqrt(number)) + 1):if int(math.sqrt(number - i**2))**2 == number - i**2:

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Question

Half code required for this question for complete.

Given an integer num_perfect_squares will return the minimum amount of perfect squares are required
to sum to the specified number. Lagrange's four-square theorem gives us that the answer will always
be between 1 and 4 (https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem).
Some examples:
Number | Perfect Squares representation | Answer
-|--
-|------
9
|3^2
10
| 3^2+1^2
12 12^2 + 2^2 + 2^2
31 | 5^2 + 2^2+1^2+1^2
| 1
return 1
12
13
14
import math
def num_perfect_squares(number):
Returns the smallest number of perfect squares that sum to the specified number.
:return: int between 1 - 4
# If the number is a perfect square then we only need 1 number.
if int(math.sqrt(number))**2 :
==
number:
#We check if https://en.wikipedia.org/wiki/Legendre%27s_three-square_theorem holds and divide
# the number accordingly. le. if the number can be written as a sum of 3 squares (where the
# 0^2 is allowed), which is possible for all numbers except those of the form: 4^a(8b + 7).
while number > 0 and number % 4 == 0:
number /= 4
# If the number is of the form: 4^a(8b + 7) it can't be expressed as a sum of three (or less
# excluding the 0^2) perfect squares. If the number was of that form, the previous while loop
# divided away the 4^a, so by now it would be of the form: 8b + 7. So check if this is the case
# and return 4 since it neccessarily must be a sum of 4 perfect squares, in accordance
# with https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem.
if number % 8 == 7:
return 4
# By now we know that the number wasn't of the form 4^a(8b + 7) so it can be expressed as a sum
# of 3 or less perfect squares. Try first to express it as a sum of 2 perfect squares, and if
# that fails, we know finally that it can be expressed as a sum of 3 perfect squares.
for i in range(1, int(math.sqrt(number)) + 1):
if int(math.sqrt(number - i**2))**2 == number - i**2:

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