SE-SKOLORNAS 1974

Let $a_n = 2^{n-1}$ for $n > 0$ . Let $$b_n = \sum\limits_{r+s \leq n} a_ra_s$$ Find $b_n-b_{n-1}$ , $b_n-2b_{n-1}$ and $b_n$ .

Show that $$1 - \frac{1}{k} \leq n\left(\sqrt[n]{k}-1\right) \leq k - 1$$ for all positive integers $n$ and positive reals $k$ .

Let $a_1=1$ , $a_2=2^{a_1}$ , $a_3=3^{a_2}$ , $a_4=4^{a_3}$ , $\dots$ , $a_9 = 9^{a_8}$ . Find the last two digits of $a_9$ .

Find all polynomials $p(x)$ such that $p(x^2) = p(x)^2$ for all $x$ . Hence find all polynomials $q(x)$ such that $$q\left(x^2 - 2x\right) = q\left(x-2\right)^2$$

Find the smallest positive real $t$ such that $$\left\{ \begin{array}{l} x_1 + x_3 = 2t x_2 x_2 + x_4 = 2t x_3 x_3 + x_5=2t x_4 \end{array} \right.$$ has a solution $x_1$ , $x_2$ , $x_3$ , $x_4$ , $x_5$ in non-negative reals, not all zero.

For which $n$ can we find positive integers $a_1,a_2,\dots,a_n$ such that $$a_1^2+a_2^2+\cdots+a_n^2$$ is a square?