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# Square pyramidal number (Sum of Squares)

• Difficulty Level : Medium
• Last Updated : 06 Apr, 2021

A Square pyramidal number represents sum of squares of first natural numbers. First few Square pyramidal numbers are 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, …
Geometrically these numbers represent number of spheres to be stacked to form a pyramid with square base. Please see this Wiki image for more clarity.
Given a number s (1 <= s <= 1000000000). If s is sum of the squares of the first n natural numbers then print n, otherwise print -1.
Examples :

```Input : 14
Output : 3
Explanation : 1*1 + 2*2 + 3*3 = 14

Input : 26
Output : -1```

A simple solution is to run through all numbers starting from 1, compute current sum. If current sum is equal to given sum, then we return true, else false.

## C++

 `// C++ program to check if a``// given number is sum of``// squares of natural numbers.``#include ``using` `namespace` `std;` `// Function to find if the``// given number is sum of``// the squares of first n``// natural numbers``int` `findS(``int` `s)``{``    ``int` `sum = 0;` `    ``// Start adding squares of``    ``// the numbers from 1``    ``for` `(``int` `n = 1; sum < s; n++)``    ``{``        ``sum += n * n;` `        ``// If sum becomes equal to s``        ``// return n``        ``if` `(sum == s)``            ``return` `n;``    ``}` `    ``return` `-1;``}` `// Drivers code``int` `main()``{``    ``int` `s = 13;``    ``int` `n = findS(s);``    ``n == -1 ? cout << ``"-1"` `: cout << n;` `    ``return` `0;``}`

## Java

 `// Java program to check if a``// given number is sum of``// squares of natural numbers.``class` `GFG``{` `    ``// Function to find if the``    ``// given number is sum of``    ``// the squares of first``    ``// n natural numbers``    ``static` `int` `findS(``int` `s)``    ``{``        ``int` `sum = ``0``;` `        ``// Start adding squares of``        ``// the numbers from 1``        ``for` `(``int` `n = ``1``; sum < s; n++)``        ``{``            ``sum += n * n;` `            ``// If sum becomes equal to s``            ``// return n``            ``if` `(sum == s)``                ``return` `n;``        ``}` `        ``return` `-``1``;``    ``}` `    ``// Drivers code``    ``public` `static` `void` `main(String[] args)``    ``{` `        ``int` `s = ``13``;``        ``int` `n = findS(s);``        ``if` `(n == -``1``)``            ``System.out.println(``"-1"``);``        ``else``            ``System.out.println(n);``    ``}``}`

## Python3

 `# Python3 program to find if``# the given number is sum of``# the squares of first``# n natural numbers` `# Function to find if the given``# number is sum of the squares``# of first n natural numbers``def` `findS (s):``    ``_sum ``=` `0``    ``n ``=` `1``    ` `    ``# Start adding squares of``    ``# the numbers from 1``    ``while``(_sum < s):``        ``_sum ``+``=` `n ``*` `n``        ``n``+``=` `1``    ``n``-``=` `1``    ` `    ``# If sum becomes equal to s``    ``# return n``    ``if` `_sum ``=``=` `s:``        ``return` `n``    ``return` `-``1` `# Driver code``s ``=` `13``n ``=` `findS (s)``if` `n ``=``=` `-``1``:``    ``print``(``"-1"``)``else``:``    ``print``(n)`

## C#

 `// C# program to check if a given``// number is sum of squares of``// natural numbers.``using` `System;` `class` `GFG``{``    ` `    ``// Function to find if the given``    ``// number is sum of the squares``    ``// of first n natural numbers``    ``static` `int` `findS(``int` `s)``    ``{``        ``int` `sum = 0;``    ` `        ``// Start adding squares of``        ``// the numbers from 1``        ``for` `(``int` `n = 1; sum < s; n++)``        ``{``            ``sum += n * n;``    ` `            ``// If sum becomes equal``            ``// to s return n``            ``if` `(sum == s)``                ``return` `n;``        ``}``    ` `        ``return` `-1;``    ``}``    ` `    ``// Drivers code``    ``public` `static` `void` `Main()``    ``{``        ``int` `s = 13;``        ` `        ``int` `n = findS(s);``        ` `        ``if``(n == -1)``            ``Console.Write(``"-1"``) ;``        ``else``            ``Console.Write(n);``    ``}``}``// This code is contribute by``// Smitha Dinesh Semwal`

## PHP

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## Javascript

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OUTPUT :

`-1`

An alternate solution is to use Newton Raphson Method
We know sum of squares of first n natural numbers is n * (n + 1) * (2*n + 1) / 6.
We can write solutions as
k * (k + 1) * (2*k + 1) / 6 = s
k * (k + 1) * (2*k + 1) – 6s = 0
We can find roots of above cubic equation using Newton Raphson Method, then check if root is integer or not.

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