# squareTerms4

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``````function divisor(n, factor) {
var divisor = 1;
while (n % factor == 0) {
n = n / factor;
divisor = divisor * factor;
}
return divisor;
}

function getPrimesUntil(n) {
// Prime sieve algorithm
var range = Math.floor(Math.sqrt(n)) + 1;
var isPrime = Array(n).fill(1);
var primes = [2];
for (var m = 3; m < range; m += 2) {
if (isPrime[m]) {
primes.push(m);
for (var k = m * m; k <= n; k += m) {
isPrime[k] = 0;
}
}
}
for (var m = range + 1 - (range % 2); m <= n; m += 2) {
if (isPrime[m]) primes.push(m);
}
return {
primes: primes,
factorize: function (n) {
var p, count, primeFactors;
// Trial division algorithm
if (n < 2) return [];
primeFactors = [];
for (p of this.primes) {
count = 0;
while (n % p == 0) {
count++;
n /= p;
}
if (count) primeFactors.push({value: p, count: count});
}
if (n > 1) {
primeFactors.push({value: n, count: 1});
}
return primeFactors;
}
}
}

function squareTerms4(n) {
var n1, n2, n3, n4, sq, sq1, sq2, sq3, sq4, primes, factors, f, f3, factors3, ok,
res1, res2, res3, res4;
primes = getPrimesUntil(n);
factors = primes.factorize(n);
res1 = n > 0 ? 1 : 0;
res2 = res3 = res4 = 0;
for (f of factors) { // For each of the factors:
n1 = f.value;
// 1. Find a suitable first square
for (sq1 = Math.floor(Math.sqrt(n1)); sq1>0; sq1--) {
n2 = n1 - sq1*sq1;
// A number can be written as a sum of three squares
// <==> it is NOT of the form 4^a(8b+7)
if ( (n2 / divisor(n2, 4)) % 8 !== 7 ) break; // found a possibility
}
// 2. Find a suitable second square
for (sq2 = Math.floor(Math.sqrt(n2)); sq2>0; sq2--) {
n3 = n2 - sq2*sq2;
// A number can be written as a sum of two squares
// <==> all its prime factors of the form 4a+3 have an even exponent
factors3 = primes.factorize(n3);
ok = true;
for (f3 of factors3) {
ok = (f3.value % 4 != 3) || (f3.count % 2 == 0);
if (!ok) break;
}
if (ok) break;
}
// To save time: extract the largest square divisor from the previous factorisation:
sq = 1;
for (f3 of factors3) {
sq *= Math.pow(f3.value, (f3.count - f3.count % 2) / 2);
f3.count = f3.count % 2;
}
n3 /= sq*sq;
// 3. Find a suitable third square
sq4 = 0;
// b. Find square for the remaining value:
for (sq3 = Math.floor(Math.sqrt(n3)); sq3>0; sq3--) {
n4 = n3 - sq3*sq3;
// See if this yields a sum of two squares:
sq4 = Math.floor(Math.sqrt(n4));
if (n4 == sq4*sq4) break; // YES!
}
// Incorporate the square divisor back into the step-3 result:
sq3 *= sq;
sq4 *= sq;
// 4. Merge this quadruple of squares with any previous
while (f.count--) {
[res1, res2, res3, res4] = [
Math.abs(res1*sq1 + res2*sq2 + res3*sq3 + res4*sq4),
Math.abs(res1*sq2 - res2*sq1 + res3*sq4 - res4*sq3),
Math.abs(res1*sq3 - res2*sq4 - res3*sq1 + res4*sq2),
Math.abs(res1*sq4 + res2*sq3 - res3*sq2 - res4*sq1)
];
}
}
// Return the 4 squares in descending order (for convenience):
return [res1, res2, res3, res4].sort( (a,b) => b-a );
}

// Produce the result for some random input number
var n = Math.floor(Math.random() * 1000000);
var solution = squareTerms4(n);
// Perform the sum of squares to see it is correct:
var check = solution.reduce( (a,b) => a+b*b, 0 );
if (check !== n) throw "FAILURE: difference " + n + " - " + check;
// Print the result
console.log(n + ' = ' + solution.map( x => x+'Β²' ).join(' + '));``````
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