Recommended root canal

Let $S_n$ be the set of roots of $f_n(x) = (1+x)^n - x^n - 1 = 0$ . If $m < n$ and $S_m = S_n$ , then $S_n$ must contain at least one repeated root. Suppose that $u \in \mathbb{C}$ is a repeated root in $S_n$ . Then $f_n(u) = (1+u)^n - u^n - 1 = 0$ and $f_n'(u) = n(1+u)^{n-1} - nu^{n-1} = 0$ , so that $1 + u^n \; = \; (1+u)^n \; = \; (1+u)u^{n-1} \; = \; u^{n-1} + u^n$ and hence $(1+u)^{n-1} = u^{n-1} = 1$ . Since $u$ and $1+u$ are $(n-1)$ st roots of unity (and hence complex numbers of modulus $1$ ) which differ by $1$ , we deduce that $u \; = \; -\tfrac12 \pm \tfrac12i\sqrt{3} \;=\; \omega,\omega^2 \qquad 1+u \; = \; \tfrac12 \pm \tfrac12i\sqrt{3} \; = \; -\omega^2,-\omega$ where $\omega$ is the primitive third root of unity. Since both $u$ and $1+u$ are $(n-1)$ st roots of unity, we deduce that $n-1$ is a multiple of $6$ .

Thus $\omega$ and $\omega^2$ are the only possible roots of $S_n$ that can be repeated. Since $f_n''(x) = n(n-1)\big[(1+x)^{n-2} - x^{n-2}\big]$ it is clear that $f_n''(\omega) \neq 0$ , and so $\omega$ cannot be more than a double root in $S_n$ (and similarly for $\omega^2$ ). Since $f_n(x)$ has real coefficients, we see that both $\omega$ and $\omega^2$ occur as double roots of $S_n$ . Thus we deduce that $S_n$ contains $n-2$ elements.

Hence we deduce that $S_m$ must have $n-2$ roots, including $\omega$ and $\omega^2$ . If these two were double roots, then $m-1$ would be a multiple of $6$ , as for $n$ , and $S_m$ would contain $m-2$ roots. The only way this would be possible would be if $m=n$ , which is not the case. This we deduce that $S_m$ contains no double roots, and hence that $m=n-2$ . But this would imply (since $f_n(x)$ has leading term $nx^{n-1}$ ) that $f_n(x) \; = \; \tfrac{n}{n-2}(x^2+x+1)f_{n-2}(x)$ Using the Binomial Theorem and multiplying out the first few terms, $\begin{array}{rcl} f_n(x) & = & \tfrac{n}{1}x^{n-1} + \tfrac{n(n-1)}{2}x^{n-2} + \tfrac{n(n-1)(n-2)}{6}x^{n-3} + \cdots \\ f_{n-2}(x) & = & \tfrac{n-2}{1}x^{n-3} + \tfrac{(n-2)(n-3)}{2}x^{n-4} + \tfrac{(n-2)(n-3)(n-4)}{6}x^{n-5} +\cdots \\ (x^2 + x + 1)f_{n-2}(x) & = & \tfrac{n-2}{1}x^{n-1} + \tfrac{(n-1)(n-2)}{2}x^{n-2} + \tfrac{(n-2)(n^2-4n+9)}{6}x^{n-3} + \cdots \end{array}$ and so, matching the coefficients of $x^{n-3}$ , $\begin{array}{rcl} \tfrac16n(n-1)(n-2) & = & \tfrac{n}{n-2} \times \tfrac16(n-2)(n^2 - 4n + 9) \\ n^2 - 3n + 2 \; = \; (n-1)(n-2) & = & n^2 - 4n + 9 \end{array}$ and hence $n=7$ . We can calculate that $(1+x)^5 - x^5 - 1 \, = \, 5x(x+1)(x^2+x+1)$ , while $(1+x)^7 - x^7 - 1 \,=\, 7x(x+1)(x^2+x+1)^2$ , and hence the case $m=5$ , $n=7$ is possible. Thus the largest, indeed the only, value of $m\times n$ where $m$ and $n$ are distinct and $S_m = S_n$ is $5\times7=35$ .

Once we have limited the possible values of double/repeated roots of these polynomials, the rest is relatively straightforward. Note that the coefficient of $x^{n-3}$ is the first one which is not automatically the same on both sides of the identity $f_n(x) = \tfrac{n}{n-2}(x^2+x+1)f_{n-2}(x)$ .

By substituting integers into n and m, we'll be able to get n=5 and m=7. Therefore the product of n and m is 35.

It's really not hard to do! You could just pull up Wolfram|Alpha, but it's not too hard to hack out. When you do so, you see that $m=7$ and $m=5$ have the same roots. I didn't check for any others, but it turns out that $\boxed{35}$ is the correct answer.