**π‘ Problem Formulation:** The task is to determine if, by simply adding 1 to a given integer, the result can become a perfect square in Python. This means that for an input number `n`

, there needs to be an integer `x`

such that `(n+1) = x^2`

. For example, given the input 8, adding 1 gives us 9, which is `3^2`

, hence the output is True. Conversely, for the input 10, adding 1 gives us 11, which is not a perfect square, hence the output is False.

## Method 1: Using the Square Root Function

An approach to determine if the incremented number is a perfect square is to use the square root function `math.sqrt()`

and check if the square root of `n+1`

is an integer. This is a straightforward method using the Python standard library.

Here’s an example:

import math def is_perfect_square_plus_one(n): return math.isqrt(n + 1)**2 == n + 1 print(is_perfect_square_plus_one(8)) print(is_perfect_square_plus_one(10))

Output:

True False

This snippet defines a function that adds 1 to the input number `n`

, takes the integer square root of the result with `math.isqrt()`

, and then squares it to check if it equals `n+1`

. This confirms if the number after adding 1 is a perfect square. Notably, `math.isqrt()`

ensures the square root result is an integer.

## Method 2: Iterative Guess and Check

One can iteratively check each number starting from 1, squaring it, until the square is equal to `n+1`

or has surpassed it. This method is simple but not efficient for large numbers.

Here’s an example:

def is_perfect_square_plus_one(n): x = 1 while x*x <= n + 1: if x*x == n + 1: return True x += 1 return False print(is_perfect_square_plus_one(8)) print(is_perfect_square_plus_one(10))

Output:

True False

In this example, the function starts with `x=1`

and incrementally checks if `x^2`

equals `n+1`

. If it finds such an `x`

, it returns True; otherwise, it continues until `x^2`

surpasses `n+1`

, then it returns False.

## Method 3: Binary Search

To improve efficiency, especially with larger numbers, one can use binary search to find if `n+1`

is a perfect square. Starting with a range that holds the potential square root, the method progressively halves the search space.

Here’s an example:

def is_perfect_square_plus_one(n): left, right = 0, n+1 while left <= right: mid = (left + right) // 2 mid_squared = mid * mid if mid_squared == n + 1: return True elif mid_squared < n + 1: left = mid + 1 else: right = mid - 1 return False print(is_perfect_square_plus_one(8)) print(is_perfect_square_plus_one(10))

Output:

True False

This function uses binary search logic to narrow down the possible square root of `n+1`

. By comparing and halving the interval, it efficiently determines if `n+1`

is a perfect square, greatly reducing the time complexity compared to linear search methods.

## Method 4: Using List Comprehension and Any

By utilizing list comprehension in combination with the `any`

function, one can test if any number squared equals `n+1`

within a specified range. It’s a more Pythonic and compact way of doing a linear search method.

Here’s an example:

def is_perfect_square_plus_one(n): return any(x * x == n + 1 for x in range(1, n + 2)) print(is_perfect_square_plus_one(8)) print(is_perfect_square_plus_one(10))

Output:

True False

This code utilizes the succinctness of list comprehensions and the `any`

function to find if there is any integer `x`

in the range from 1 to `n+1`

such that `x^2`

equals `n+1`

. Although it’s more elegant, it has similar performance characteristics to the iterative guess and check method.

## Bonus One-Liner Method 5: Using Power and Modulo Operator

A concise one-liner method involves incrementing the number by 1 and checking if `(n+1)**0.5 % 1 == 0`

, which asserts that the square root is an integer.

Here’s an example:

is_perfect_square_plus_one = lambda n: (n + 1)**0.5 % 1 == 0 print(is_perfect_square_plus_one(8)) print(is_perfect_square_plus_one(10))

Output:

True False

The lambda function checks whether the square root of `n+1`

is an integer by using the modulo operator on `(n+1)**0.5`

. If there’s no remainder, then the number is a perfect square. This method is the most concise but could run into floating-point arithmetic issues for very large numbers.

## Summary/Discussion

**Method 1: Using the Square Root Function.**Straightforward and efficient for all ranges of integers. Utilizes Python’s built-in math library functions.**Method 2: Iterative Guess and Check.**Simple to understand. However, it is inefficient for large numbers as it involves a linear search.**Method 3: Binary Search.**Efficient for large numbers. Reduces computation time by continually halving the search space. Preferred for high-performance needs.**Method 4: Using List Comprehension and Any.**Elegant and Pythonic. Offers no performance benefits over the linear search but is more readable.**Method 5: Using Power and Modulo Operator.**Most concise option. Good for quick checks and small numbers but potentially inaccurate for very large numbers due to floating-point errors.

Emily Rosemary Collins is a tech enthusiast with a strong background in computer science, always staying up-to-date with the latest trends and innovations. Apart from her love for technology, Emily enjoys exploring the great outdoors, participating in local community events, and dedicating her free time to painting and photography. Her interests and passion for personal growth make her an engaging conversationalist and a reliable source of knowledge in the ever-evolving world of technology.