# Integers have sum of squared divisors as perfect square in JavaScript

## Problem

We are required to write a JavaScript function that takes in a range specified by an array of two numbers m and n.

Our function is supposed to find all integers between m and n (m and n integers such as 1 <= m <= n) such that the sum of their squared divisors is itself a square.

It should return an array of subarrays. The subarrays will have two elements: first the number the squared divisors of which is a square and then the sum of the squared divisors.

## Example

Following is the code −

Live Demo

const range = [1, 500];
const listSquared = ([m, n]) => {
const res = [];
for (let i = m; i <= n; ++i) {
let sum = getDivisors(i).reduce((sum, n) => sum + n * n, 0);
let ok = Number.isInteger(Math.sqrt(sum));
if (ok) {
res.push([i, sum]);
}
}
return res;
}
function getDivisors (n) {
const divisors = [];
for (let i = 1; i <= n / 2; ++i) {
if (n % i) {
continue;
}
divisors.push(i);
}
return divisors.concat([n]);
}
console.log(listSquared(range));

## Output

[ [ 1, 1 ], [ 42, 2500 ], [ 246, 84100 ], [ 287, 84100 ] ]