Brahmagupta Fibonacci Identity
Last Updated :
09 Nov, 2022
Brahmagupta Fibonacci identity states that the product of two numbers each of which is a sum of 2 squares can be represented as sum of 2 squares in 2 different forms.
Mathematically,
If a = p^2 + q^2 and b = r^2 + s^2
then a * b can be written in two
different forms:
= (p^2 + q^2) * (r^2 + s^2)
= (pr – qs)^2 + (ps + qr)^2 …………(1)
= (pr + qs)^2 + (ps – qr)^2 …………(2)
Some Examples are:
a = 5(= 1^2 + 2^2)
b = 25(= 3^2 + 4^2)
a*b = 125
Representation of a * b as sum of 2 squares:
2^2 + 11^2 = 125
5^2 + 10^2 = 125
Explanations:
a = 5 and b = 25 each can be expressed as a sum of 2 squares and their product a*b which is 125 can be expressed as sum of 2 squares in two different forms. This is according to
the Brahmagupta Fibonacci Identity and satisfies the identity condition.
a = 13(= 2^2 + 3^2)
b = 41(= 4^2 + 5^2)
a*b = 533
Representation of a * b as sum of 2 squares:
2^2 + 23^2 = 533
7^2 + 22^2 = 533
a = 85(= 6^2 + 7^2)
b = 41(= 4^2 + 5^2)
a*b = 3485
Representation of a * b as sum of 2 squares:
2^2 + 59^2 = 3485
11^2 + 58^2 = 3485
26^2 + 53^2 = 3485
37^2 + 46^2 = 3485
Below is a program to verify Brahmagupta Fibonacci identity for given two numbers which are sums of two squares.
C++
#include <bits/stdc++.h>
using namespace std;
void find_sum_of_two_squares(int a,
int b)
{
int ab = a*b;
for (int i = 0; i * i <= ab; i++)
{
for (int j = i; i * i +
j * j <= ab; j++)
{
if (i * i + j * j == ab)
cout << i << "^2 + " << j
<< "^2 = " << ab << "\n";
}
}
}
int main()
{
int a = 1 * 1 + 2 * 2;
int b = 3 * 3 + 4 * 4;
cout << "Representation of a * b as sum"
" of 2 squares:\n";
find_sum_of_two_squares(a, b);
}
|
Java
class GFG
{
static void find_sum_of_two_squares(int a,
int b)
{
int ab = a * b;
for (int i = 0; i * i <= ab; i++)
{
for (int j = i; i * i +
j * j <= ab; j++)
{
if (i * i + j * j == ab)
System.out.println(i + "^2 + " +
j +"^2 = " + ab);
}
}
}
public static void main(String[] args)
{
int a = 1 * 1 + 2 * 2;
int b = 3 * 3 + 4 * 4;
System.out.println("Representation of a * b " +
"as sum of 2 squares:");
find_sum_of_two_squares(a, b);
}
}
|
Python 3
def find_sum_of_two_squares(a, b):
ab = a * b
i=0;
while(i * i <= ab):
j = i
while(i * i + j * j <= ab):
if (i * i + j * j == ab):
print(i,"^2 + ",j,"^2 = ",ab)
j += 1
i += 1
a = 1 * 1 + 2 * 2
b = 3 * 3 + 4 * 4
print("Representation of a * b as sum"
" of 2 squares:")
find_sum_of_two_squares(a, b)
|
C#
using System;
class GFG
{
static void find_sum_of_two_squares(int a,
int b)
{
int ab = a * b;
for (int i = 0; i * i <= ab; i++)
{
for (int j = i; i * i +
j * j <= ab; j++)
{
if (i * i + j * j == ab)
Console.Write(i + "^2 + " + j +
"^2 = " + ab + "\n");
}
}
}
public static void Main()
{
int a = 1 * 1 + 2 * 2;
int b = 3 * 3 + 4 * 4;
Console.Write("Representation of a * b " +
"as sum of 2 squares:\n");
find_sum_of_two_squares(a, b);
}
}
|
PHP
<?php
function find_sum_of_two_squares($a, $b)
{
$ab = $a * $b;
for ($i = 0; $i * $i <= $ab; $i++)
{
for ($j = $i; $i * $i +
$j * $j <= $ab; $j++)
{
if ($i * $i + $j * $j == $ab)
echo $i ,"^2 + ", $j ,
"^2 = " , $ab ,"\n";
}
}
}
$a = 1 * 1 + 2 * 2;
$b = 3 * 3 + 4 * 4;
echo "Representation of a * b ".
"as sum of 2 squares:\n";
find_sum_of_two_squares($a, $b);
?>
|
Javascript
<script>
function find_sum_of_two_squares(a, b)
{
let ab = a * b;
for (let i = 0; i * i <= ab; i++)
{
for (let j = i; i * i +
j * j <= ab; j++)
{
if (i * i + j * j == ab)
document.write(i + "^2 + " +
j +"^2 = " + ab + "<br/>");
}
}
}
let a = 1 * 1 + 2 * 2;
let b = 3 * 3 + 4 * 4;
document.write("Representation of a * b " +
"as sum of 2 squares:" + "<br/>");
find_sum_of_two_squares(a, b);
</script>
|
Output :
Representation of a * b as sum of 2 squares:
2^2 + 11^2 = 125
5^2 + 10^2 = 125
Time complexity : O(a*b)
Auxiliary Space : O(1)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...