Huge linear system of equations with powers of $2$

Question

I've got a large system of equations: $$ \begin{align*} (2^0)^na_n + (2^0)^{n-1}a_{n-1} + \cdots + (2^0)^1a_1 &= 4^0 \\ (2^1)^na_n + (2^1)^{n-1}a_{n-1} + \cdots + (2^1)^1a_1 &= 4^1 \\ \vdots\\ (2^{n-1})^na_n + (2^{n-1})^{n-1}a_{n-1} + \cdots + (2^{n-1})^1a_1 &= 4^{n-1} \\ \end{align*} $$ for $n\geq2$. I'd like to show that the (unique) solution of this system is when $a_2=1$ and all other variables are zero. I'm not very familiar with linear algebra, however I tried putting these equations in matrix form. My attempt involved using induction of variable $n$, and thinking of one matrix as a "sub-matrix" of the next one. Again, due to limited knowledge of linear algebra, I got nowhere with it.

Also, is induction a good idea for such a problem or would you use another method of proof?

Thank you.

Answer 1

Suppose that $(a_1,\ldots,a_n)$ is a solution of the proposed system. Consider $P(X)=-X^2+\sum_1^na_jX^j$. This is a polynomial of degree smaller or equal to $n$, with $P(0)=0$. Moreover, by assumption $P(2^{i})=0$ for every $i=0,\ldots n-1$, so $P$ has at $n+1$ zeros, and it must be identically zero. So $(a_1,a_2,\ldots,a_n)=(0,1,\ldots,0)$. The converse is trivially true.

Answer 2

Use Cramer's rule remembering the $4^k=(2^k)^2$ which makes the right hand side the same as the second to last column. This means that two columns will be the same (and so the determinant will be zero) for all variables except for $a_2$ where the determinant in the numerator will be the same as the determinant in the denominator. Cramer's rule can be found here: http://en.wikipedia.org/wiki/Cramer's_rule