$(CZS 6)$ Let $d$ and $p$ be two real numbers. Find the first term of an arithmetic progression $a_1, a_2, a_3, \cdots$ with difference $d$ such that $a_1a_2a_3a_4 = p.$ Find the number of solutions in terms of $d$ and $p.$
$ a(a+d)(a+2d)(a+3d)=(a^{2}+3ad)(a^{2}+3ad+2d^{2}) $.
note $ x=a^{2}+3ad $ and now we only have to solve the quadratic equation in $ x $.
for find the number of solutions we only have to analyze the sign of the discriminant.
in this way we'll obtain $ x $ like a function in $ d,p $ and we'll make another equation in $ a $ and again we analyze the sign of this new discriminant.