x²+px+q = 0

The solutions are

 and 


We know that the solutions are rational, so the
discriminant p²-4q is the square of an integer n,
then

     p²-4q = n²        and the solutions are
     p²-n² = 4q         and 
(p-n)(p+n) = 4q

1. The right side 4q is even so the left side (p-n)(p+n) is even.
2. The sum or difference of two even integers is even.
3. The sum or difference of two odd integers is even.
4. The sum or difference of an even and an odd integer is odd.
5. The product of two even integers is even.
6. The product of two odd integers is odd.
7. If an integer is odd, so is its opposite.
8. An even integer divided by 2 is an integer.

Case 1. p is even and n is odd
Then by 4, p-n and p+n are both odd.
Then by 6, (p-n)(p+n) is odd.
That contradicts 1.  Thus case 1 is ruled out.

Case 2. p is odd and n is even
Then by 4, p-n and p+n are both odd.
Then by 6, (p-n)(p+n) is odd.
That contradicts 2.  Thus case 2 is ruled out.

Case 3. p and n are both even
Then because of 2 and 7,
-p+n and -p-n are both even
Therefore, by 8,  and  are both
integers.  The proof is complete for case 3.

Case 4. p and n are both odd
Then because of 3 and 7
-p+n and -p-n are both even
Therefore, by 8,  and  are both
integers.  The proof is complete for case 4.

Edwin