Performing power operations on an array of numbers in JavaScript

We need to write a JavaScript function that takes an array of integers with even length and performs a specific power operation to find two numbers whose squares sum to a calculated value.

Problem Statement

Given an array of integers with even length, we calculate:

num = (arr[0]² + arr[1]²) × (arr[2]² + arr[3]²) × ... × (arr[n-2]² + arr[n-1]²)

Our function should return an array [A, B] such that A² + B² = num.

Example Walkthrough

For array [1, 2, 3, 4]:

• First pair: 1² + 2² = 1 + 4 = 5
• Second pair: 3² + 4² = 9 + 16 = 25
• num = 5 × 25 = 125
• Find A, B where A² + B² = 125
• Result: [2, 11] because 2² + 11² = 4 + 121 = 125

Solution

const arr = [1, 2, 3, 4];

const findMatchingSumArray = (arr = []) => {
    let squaredSum = 1;
    
    // Calculate the product of sum of squares for each pair
    for(let i = 0; i 

[2, 11]
Verification: 125


How It Works

The algorithm works in two phases:



Calculate the target sum: Iterate through pairs of array elements, square each element, sum the squares in each pair, then multiply all pair sums together.

Find the solution: Use nested loops to test all combinations of integers k and j until we find k² + j² equals our target sum.


Testing with Different Arrays

// Test with different arrays
const testCases = [
    [1, 2, 3, 4],
    [1, 1, 1, 1], 
    [2, 3, 4, 5]
];

testCases.forEach(testArr => {
    const result = findMatchingSumArray(testArr);
    console.log(`Array: [${testArr.join(', ')}] => Result: [${result.join(', ')}]`);
});


Array: [1, 2, 3, 4] => Result: [2, 11]
Array: [1, 1, 1, 1] => Result: [0, 2]
Array: [2, 3, 4, 5] => Result: [1, 18]


Conclusion

This algorithm efficiently solves the power operation problem by first calculating the target sum from paired squares, then using a brute-force search to find two integers whose squares sum to that target. The nested loop approach ensures we find the solution systematically.