Factoring

Given $a^2 + b^2 + cab$ for some real number $c \in (-2, 2)$ , we will factor it to $(a + b + p\sqrt{ab})(a + b + q\sqrt{ab})$ for some real numbers $p, q$ . We will determine what $p, q$ should be.

If we multiply out the factorization, we obtain

$\begin{aligned} (a + b + p\sqrt{ab})(a + b + q\sqrt{ab}) &= (a+b)^2 + (p+q) \cdot (a+b) \sqrt{ab} + pq \cdot ab \\ &= a^2 + 2ab + b^2 + (p+q) \cdot (a+b) \sqrt{ab} + pq \cdot ab \\ &= a^2 + b^2 + (2+pq) \cdot ab + (p+q) \cdot (a+b) \sqrt{ab} \end{aligned}$

Thus, if we want this to be equal to $a^2 + b^2 + cab$ , we need $2+pq = c$ and $p+q = 0$ . The latter means $q = -p$ ; substituting to the former gives $2 - p^2 = c$ or $p = \pm \sqrt{2-c}$ . Without loss of generality, let $p$ be the positive value $\sqrt{2-c}$ , so $q$ gets the negative value $-\sqrt{2-c}$ .

First, notice that since we assume $c < 2$ , the square root is defined, so $p,q$ are defined. Next, since we also assume $c > -2$ , we have $p < 2$ (and thus $q > -2$ ). Also, clearly $0 < p$ and $0 > q$ . Thus $p,q \in (-2, 2)$ .

The important point is that if we have a factor $x + y + c \sqrt{xy}$ , and we substitute $x = a^2, y = b^2$ and factorize it further to $(a + b + p \sqrt{ab})(a + b + q \sqrt{ab}$ , the following fact holds: if $c \in (-2, 2)$ , then $p,q \in (-2, 2)$ as well. Thus if we factorize each of its factors and so on, all coefficients will stay being real numbers.

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Now we're ready for the real problem. However, before computing $f(k)$ , we will first compute $g(k) = \min_{1 \le i \le 2^k} c_i$ . We claim that $g(1) = -\sqrt{2}$ and $g(k+1) = - \sqrt{2 - g(k)}$ . The first claim is easy; just plug in $c = 0$ to the formula $q = -\sqrt{2-c}$ above to get $q = -\sqrt{2}$ . Since $p > 0$ , we have $g(1) = q = -\sqrt{2}$ .

To prove the recurrence, consider $x^{2^{k+1}} + y^{2^{k+1}}$ . Substitute $t = x^2, u = y^2$ , so our expression becomes $t^{2^k} + u^{2^k}$ . We can factorize this to have the factor $t + u + g(k) \sqrt{tu} = x^2 + y^2 + g(k) xy$ . Plugging this to the above formula, we obtain that this can be further factorized into $(x + y + \sqrt{2 - g(k)} \cdot \sqrt{xy})(x + y - \sqrt{2 - g(k)} \cdot \sqrt{xy})$ , so we see $g(k+1) \le -\sqrt{2-g(k)}$ .

To see that it is indeed the minimum, consider any factor $t + u + c \sqrt{tu}$ . By definition, $c \ge g(k)$ . Thus if we factorize $x^2 + y^2 + cxy = (x + y + p \sqrt{xy})(x + y + q \sqrt{xy})$ , we obtain $q = -\sqrt{2-c} \ge -\sqrt{2-g(k)}$ . (And clearly $p > 0$ .) So in fact $g(k+1) = -\sqrt{2-g(k)}$ ; all factors of $x^{2^{k+1}} + y^{2^{k+1}}$ have those coefficients at least $-\sqrt{2-g(k)}$ .

Finally, we will now go to $f(k)$ . A very similar argument as above gives that $f(k+1) = \sqrt{2 - g(k)}$ : this is caused from taking the positive coefficient of factoring $x^2 + y^2 + g(k) xy$ , and any other factor $x^2 + y^2 + cxy$ will give $p = \sqrt{2-c} \le \sqrt{2-g(k)}$ . This also shows that $f(k) = -g(k)$ , so we obtain the recurrence $f(k+1) = \sqrt{2 + f(k)}$ (with starting value $f(1) = \sqrt{2}$ ).

Finally, we're looking for the limit, so it will be in the well-known form $\sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}} = \boxed{2}$ .

This problem can be generalised to any positive integer $n$ by

$\displaystyle \prod _{k=1}^n\left(x+y+2\cos \left(\frac{\left(2k-1\right)\pi }{2n}\right)\sqrt{xy}\right)=x^n + y^n$

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To find that, notice that for a fixed $y$ , $x^n + y^n=0$ must have the same roots as $(x + y + c_1 \sqrt{xy}) (x + y + c_2 \sqrt{xy}) \ldots (x + y + c_n \sqrt{xy})=\prod_{k=1}^{n}(x+y+c_k\sqrt{xy})=0$

Now, $x^n + y^n=0$ has roots $x = ye^{i(2k-1)\pi / n}$ for $1 \le k \le n$ . Since the product above has exactly $n$ terms, which is equal to the number of roots for $x^n + y^n=0$ , we can conclude that for every $1 \le k \le n$ , there must exist a term in the product that equates to $0$ when $x = ye^{i(2k-1)\pi / n}$ . Substituting the roots into the product yields:

$\prod_{k=1}^{n}ye^{i(2k-1)\pi / n}+y+c_k\sqrt{ye^{i(2k-1)\pi / n}y} = \prod_{k=1}^{n} y\left(e^{i(2k-1)\pi / n}+1+c_ke^{i(2k-1)\pi / 2n}\right)=0, \quad [1 \le k \le n ]$

From here we can solve for $c_k$ :

$e^{i(2k-1)\pi / n}+1+c_ke^{i(2k-1)\pi / 2n}=0$

$c_{k} = -2\cos \left(\frac{\left(2k-1\right)\pi }{2n}\right), {1 \le k \le n}$

The final answer of $2$ follows easily from here by considering $k=n$