RMO 1995

The simple way to do this is to analyse the discriminant which i think is too elementary to be discussed here. Second, I'm still not out of the shock that this was an RMO problem

Easier solution using some competitive knowledge and the quadratic formula: By the quadratic formula, the roots are in the form \dfrac{-7 \pm \sqrt{49+56(q^2+1)}{2} . We know that every perfect square must be either equivalent to $0$ or $1$ \text{(mod 4)\}. \(49 \equiv 3 \text{ (mod 4)} , and any multiple of $56$ is equivalent to $0 \text{ (mod 4)}$ . Therefore, the radicand is equivalent to $3 \text{ (mod 4)}$ and is not a perfect square. Therefore, there are no integer roots.

x^2 +7x= 14(q^2-1); 7|x; 49|x^2+7x; 49|14(q^2-1); 7|q^2-1; q^2=1(mod 7), which is impossible. Hence, the equation has no integral root. (Here ';' is used for 'implies').

Put q=0,1,.. In each case no integral roots. can't be an RMO problem.

To be an integer one of the most important condition that it has to satisfy is that,that the discriminant should be a perfect square of an integer....Now in order to do that after checking the discriminant we get 56q^2+105 ,which should be a perfect square of an integer ....Let us say it is equal to p^2 where p is an integer....Now 56q^2+105=p^2 take mod 3 ...Now it forces p and q to be of the form 3k where k is an integer.. .Now substituting the value q as 3k and p as 3s we get 56×9k^2+105=9s^2...Cancelling 3 we get 56×3k^2+35=3s^2....Now the lhs is not a multiple of 3 but the rhs is a multiple of 3 so contradiction......