## 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

Difficulty: introductory

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):
    """Return True if given number is pronic, else False."""
    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) ** .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


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