Where is the mistake in this apparently-O(n lg n) multiplication algorithm?

A recent puzzle blog post about finding three evenly spaced ones lead me to a stackoverflow question with a top answer that claims to do it in O(n lg n) time. The interesting part is that the solution involves squaring a polynomial, referencing a paper that describes how to do it in O(n lg n) time.

Now, multiplying polynomials is practically the same as multiplying numbers. The only real difference is the lack of carries. But... the carries can also be done in O(n lg n) time. For example:

var value = 100; // = 0b1100100

    var inputBitCount = value.BitCount(); // 7 (because 2^7 > 100 >= 2^6)
    var n = inputBitCount * 2; // 14
    var lgn = n.BitCount(); // 4 (because 2^4 > 14 => 2^3)
    var c = lgn + 1; //5; enough space for 2n carries without overflowing

    // do apparently O(n log n) polynomial multiplication
    var p = ToPolynomialWhereBitsAreCoefficients(value); // x^6 + x^5 + x^2
    var p2 = SquarePolynomialInNLogNUsingFFT(p); // x^12 + 2x^11 + 2x^10 + x^8 + 2x^7 + x^4
    var s = CoefficientsOfPolynomial(p2); // [0,0,0,0,1,0,0,2,1,0,2,2,1]
    // note: s takes O(n lg n) space to store (each value requires at most c-1 bits)

    // propagate carries in O(n c) = O(n lg n) time
    for (var i = 0; i < n; i++)
        for (var j = 1; j < c; j++)
            if (s[i].Bit(j))
                s[i + j].IncrementInPlace();

    // extract bits of result (in little endian order)
    var r = new bool[n];
    for (var i = 0; i < n; i++)
        r[i] = s[i].Bit(0);

    // r encodes 0b10011100010000 = 10000

So my question is this: where's the mistake, here? Multiplying numbers in O(n lg n) is a gigantic open problem in computer science, and I really really doubt the answer would be this simple.

• Is the carrying wrong, or not O(n lg n)? I've worked out that lg n + 1 bits per value is enough to track the carries, and the algorithm is so simple I'd be surprised if it was wrong. Note that, although an individual increment can take O(lg n) time, the aggreg

Your algorithm is very similar to Schönhage–Strassen. In the FFT step, there are large numbers involved - as you mention, their size can be up to $O(\log n)$. Arithmetic on them doesn't come for free. You need to apply the construction recursively, and you'll lose something.

StackExchange Computer Science Q#9034, answer score: 14