Minimum 11 Multiples

Setting $f(0)$ yields that $g$ is divisible by 11. Considering $f(1)$ yields $11\mid a+b+c+d+e+f \quad (1).$ Considering $f(-1)$ gives $11\mid a-b+c-d+e-f.$ Hence with (1), we get $11\mid a+c+e$ and $11\mid b+d+f \quad$ (2). Similarly, adding up $f(3)+f(-3)$ and $f(4)+f(-4)$ yields that $0\equiv 2(3^{6}a+3^{4}c+3^{9}e)\equiv 2(3a+4c+9e)\pmod {11} \quad (3)$ and $0\equiv 2(4^{6}a+4^{4}c+4^{2}e)\equiv 2(4a+3c+5e)\pmod {11} \quad (4)$ . From here, we have $(3a+4c+9e) - 3(a+c+e) \equiv 0 \pmod{11} \Rightarrow c + 6e \equiv 0 \pmod{11}$ , and $(4a+3c+5e) - 4(a+c+e) \equiv 0 \pmod{11} \Rightarrow -c + e\equiv 0 \pmod{11}$ . Hence, $e \equiv 0 \pmod{11}$ and $c \equiv 0 \pmod{11}$ thus $a \equiv 0 \pmod{11}$ .

Now, considering $f(3) - f(-3)$ and $f(4) - f(-4)$ , we get the same equations for $b , d, f$ . Hence, we can also conclude that $b \equiv d \equiv f \equiv 0 \pmod{11}$ . So all $a, b, c, d, e, f, g$ are divisible by 11, the answer is 7.

[Note: It is preferable to stick to the same notation throughout the solution. I'm not certain why he didn't look at $f(2) + f(-2)$ , as that would seem to be the next logical choice.]

[Latex edits. Edits for clarity - Calvin]

Put x=0; you get f(x)=g, which should be a multiple of 11. Now, focus on the rest; for simplicity forget g. Put x=1; you get a+b+c+d+e+f to be a multiple of 11, and if x= -1, then a-b+c-d+e-f is 11's multiple. Then the sum of the 2 equations should also be a multiple. Add them. You get 2(a+c+e) to be a multiple of 11. Thus, if a+b+c+d+e+f is a multiple of 11 and also is a+c+e, then the rest should also be a multiple of 11, i.e. b+d+f is also a multiple of 11. Now observe... There is an expression of 7 terms, which is always a multiple of 11. I showed that one of the terms is 11's multiple and also 2 other sets each of 3 terms. In the same way it can be showed by reducing each of the 3 termed expressions (a+c+e) and (b+d+f) (which are 11's multiple); that each of the individual terms are multiples of 11.

Solution 1: Substitute in $x=0, 1, -1, 2, -2, 3, -3$ , we know that $11 \mid g, 11 \mid a+b+c+d+e+f+g, 11\mid a-b+c-d+e-f+g$ , $11 \mid 64a + 32b + 16 c + 8d +4e +2f +g, 11 \mid 64a - 32b + 16c - 8 d + 4e - 2f+g$ , $11 \mid 729a +243b +81c +27d +9e +3f +g, 11 \mid 729a -243b +81c -27d +9e -3f +g$ .

Take the sum and difference of the 2nd and 3rd, 4th and 5th, 6th and 7th equations, and dividing by out by constants coprime to 11, we get $11 \mid g, 11 \mid a + c + e, 11 \mid b + d + f$ , $11 \mid 16a + 4c + e, 11\mid 16b +4d +e$ , $11 \mid 81a +9c +e, 11 \mid 81b +9d +f$ .

Again, taking the difference between 4th and 2nd, 6th and 2nd, we get $11 \mid 5a + c, 11 \mid 10a + c$ . Hence, this gives that $11 \mid c$ , and so $11 \mid a$ , and so $11 \mid e$ . Similarly, for the other equations, we can conclude that $11 \mid d$ , $11 \mid b$ , and $11 \mid f$ . Hence, all $7$ of the coefficients must be multiples of $11$ .

Solution 2: Work modulo 11. Consider the matrix equation

$\begin{pmatrix} 1^7 & 1^6 & \ldots & 1^1 & 1 \\ 2^7 & 2^6 & \ldots & 2^1 & 1 \\ \vdots & \vdots & & \vdots & \vdots\\ 6^7 & 6^6 & \ldots & 6^1 & 1 \\ 7^7 & 7^6 & \ldots & 7^1 & 1 \\ \end{pmatrix}$ $\begin{pmatrix} a\\ b\\ \vdots\\ f\\ g\\ \end{pmatrix}$ $= \begin{pmatrix} 0\\ 0\\ \vdots\\ 0\\ 0\\ \end{pmatrix}$

The square matrix has determinant $\prod_{1 \leq i < j \leq 7} (i-j)$ , which is non-zero (modulo 11), so we can premultiply by it's inverse, to get that

$\begin{pmatrix} a\\ b\\ \vdots\\ f\\ g\\ \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ \vdots\\ 0\\ 0\\ \end{pmatrix}$

If we assume the all the coefficients are nonzero mod 11, then the congruence may have at most 6 roots in Z/11Z.But it has all 11 numbers from 0 to 10 as roots, a contradiction.Thus all the coefficients must be zero mod 11. In fact, if a polynomial of degree less than p has p zeros, then it must vanish in Z/pZ.If such a polynomial had every number as a root except 0, it must be congruent to $a(x^{p-1}-1)$ mod p, i.e. finite were primitive, it must have p-2 coefficients 0 mod p, one 1 mod p and one -1 mod p/