# 5 Best Ways to Check Whether a Number Can Be a Perfect Square After Adding 1 in Python

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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.