THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

a) We exhibit two examples. Let $f_i={\prod }_{j\ne i}(X-a_j)$, $i=1,\dots ,n$. Since the $f_i$ never vanish simultaneously, the Lagrange type interpolation polynomial $f={\sum }_{i=1}^n\frac{b_i{f}_i^2}{f_i(a_i{)}^2}$ clearly satisfies the required conditions.

Another example may be obtained by adding a suitable positive constant to the square of a suitable Lagrange interpolation polynomial; for instance, let $b$ be a positive real number less than each $b_i$, and let $$f=b+{\left(\sum _{i=1}^n\frac{\sqrt{b_i-b}}{f_i(a_i)}{f}_i\right)}^2.$$

b) Proceed by induction on $n$. The base case $n=1$ being clear, let $n\ge 2$. If some $b_i=0$, fix such an index $i$, and apply the induction hypothesis to provide a polynomial $g$ whose roots are all real, and $g(a_j)=\frac{b_j}{a_j-a_i}$ for all $j\ne i$. The polynomial $f=(X-a_i)g$ clearly satisfies the required conditions.

If no $b_i=0$, assume, without loss of generality, that $a_1<\cdots <a_n$, and let $I$ be the set of all indices $i$ such that $b_i{b}_{i+1}>0$. For each $i$ in $I$, choose $a_i^{\prime }$ and $b_i^{\prime }$ such that $a_i<a_i^{\prime }<a_{i+1}$ and $b_i{b}_i^{\prime }<0$. The roots of the Lagrange polynomial solving the interpolation problem $f(a_i)=b_i$, $i=1,\dots ,n$, and $f(a_i^{\prime })=b_i^{\prime }$, $i\in I$, are then all real and the conclusion follows.