G7.1 If the two distinct roots of the equation $a x^{2}-m x+1996=0$ are primes, find the value of $a$.
Let the roots be $\alpha, \beta \cdot \alpha+\beta=\frac{m}{a}, \alpha \beta=\frac{1996}{a}$ $1996=4 \times 499$ and 499 is a prime
$$
\therefore a=1,2,4,499,998 \text { or } 1996
$$

When $a=1, \alpha \beta=1996$, which cannot be expressed as a product of two primes $\therefore$ rejected
When $a=2, \alpha \beta=998 ; \alpha=2, \beta=499$ (accepted)
When $a=4, \alpha \beta=499$, which cannot be expressed as a product of two primes $\therefore$ rejected
When $a=499, \alpha \beta=4, \alpha=2, \beta=2$ (not distinct roots, rejected)
When $a=998, \alpha \beta=2$, which cannot be expressed as a product of two primes $\therefore$ rejected
When $a=1996, \alpha \beta=1$, which cannot be expressed as a product of two primes $\therefore$ rejected
Remark: the original question is:
If the two roots of the equation $a x^{2}-m x+1996=0$ are primes, find the value of $a$. $a=2$ or 499 (Not unique solution)