Solution for b): For a solution we will use a). Say a pair of numbers $x,y$ is good if their product is a perfect square.
From $49$ numbers take any $17$ numbers, then we have a good pair $a_1,a_2$ among them.
Now from the rest of the $47$ numbers take any $17$ numbers, then we have a good pair $a_3,a_4$ among them.
Now from the rest of the $45$ numbers take any $17$ numbers, then we have a good pair $a_5,a_6$ among them.
and so on. We repeat this process until we get last good pair $a_{33},a_{34}$ (and we are left wit $15$ numbers so that we can't repeat the process).
Now calculate their products $b_i:=a_{2i-1}a_{2i}$. So we have $17$ perfect squares $b_1,b_2,...b_{17}$. Each $b_i$ we can write like this
$$b_i = 4^x 9^y 25^z 49^t$$
Again we assign to each $b_i$ $4$-tuple $(x',y',z',t')$ where $w'$ is remainder of $w$ modulo $2$. Again two of the $b_i$ must have the same $4$-tuple, since we have $17$ numbers and only $16$ $4$-tuples. So, their product is perfect $4$-power.
Best Answer
If $b^2-4ac$ was a perfect square then the polynomial $ax^2+bx+c$ would have some rationals $\frac {p_1}{q_1}, \frac {p_2}{q_2}$ as roots($\frac{p_i}{q_i}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$). Therefore $(q_1x-p_1)(q_2x-p_2)=ax^2+bx+c$.
So $q_1,q_2,p_1,p_2$ are odd integers (since $q_1q_2=a,p_1p_2=c$) and $q_1p_2+q_2p_1=-b\Rightarrow\Leftarrow.$