## Task:
You have to create a function `isPronic` to check whether the argument passed is a Pronic Number and return true if it is & false otherwise.

### Description:

`Pronic Number` -A pronic number, oblong number, rectangular number or heteromecic number, is a number which is the product of two consecutive integers, that is, n(n + 1).

> The first few Pronic Numbers are - 0, 2, 6, 12, 20, 30, 42...

### Explanation:
    
      0 = 0 × 1   // ∴  0 is a Pronic Number
      2 = 1 × 2   // ∴  2 is a Pronic Number
      6 = 2 × 3   // ∴  6 is a Pronic Number
     12 = 3 × 4   // ∴ 12 is a Pronic Number
     20 = 4 × 5   // ∴ 20 is a Pronic Number
     30 = 5 × 6   // ∴ 30 is a Pronic Number
     42 = 6 × 7   // ∴ 42 is a Pronic Number

Source: codewars | Difficulty: EASY
Tags: Mathematics

Starter code: def is_pronic(n):
	

Solutions:
import math

def is_pronic(n):
	return n >= 0 and math.sqrt(1 + 4 * n) % 1 == 0


import math

def is_pronic(n):
	if n == 0:
		return True
	if n < 0:
		return False
	first_int = int(math.sqrt(n))
	second_int = first_int + 1
	return first_int * second_int == n


def is_pronic(n):
	if n < 0:
		return False
	s = int(n ** 0.5)
	return s * (s + 1) == n


is_pronic = lambda n: n >= 0 and (1 + 4 * n) ** 0.5 % 1 == 0


def is_pronic(n):
	return n >= 0 and int(n ** 0.5) * (int(n ** 0.5) + 1) == n


pronics = {i * (i + 1) for i in range(1000)}

def is_pronic(n):
	return n in pronics


import math

def is_pronic(n):
	if n < 0:
		return False
	a = int(math.sqrt(n))
	return a * (a + 1) == n