# Square pyramidal number (Sum of Squares)

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

 ` `

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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