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Wow, what an interesting question. I'm surprised no one has solved it yet, but that may be a consequence of the fact that no one has understood it yet. Clearly, schools need to focus more on reading comprehension ^_^.

To answer your question - not only can we prove that there exist integers a and b satisfying your assumptions, but we can even show a method of finding them. The proof is long, so sit back, relax, grab your favorite drink, and prepare to read.

First, consider that any solutions of these equations must obey the quadratic formula: x=(-b±√(b²-4ac))/2a. Substituting the coefficients of the actual equations in, we find that the solutions of the first equation are (-p±√(p²-4q))/2 and the solutions of the second equation are (-p±√(p²+4q))/2. Let us consider only the greater of the solutions in each pair - that is, replace each ± with a +. The difference between these solutions is then (-p+√(p²-4q))/2 - (-p+√(p²+4q))/2. Call this number K. We can then perform some simple algebra as follows:

(-p+√(p²-4q))/2 - (-p+√(p²+4q))/2=K

multiply both sides by 2 and simplify

√(p²-4q)-√(p²+4q)=2K

square both sides

(p²-4q) - 2√((p²-4q)(p²+4q)) + (p²+4q)=4K²

simplify

2p²-2√(p^4-16q²)=4K²

divide both sides by 2

p²-√(p^4-16q²)=2K²

subtract 2K² and add √(p^4-16q²) to both sides

p²-2K²=√(p^4-16q²)

square both sides

p^4-4K²p²+4K^4=p^4-16q²

multiply both sides by -1 and simplify

16q²=4K²p²-4K^4

divide both sides by 4

4q²=K²p²-K^4

add K^4 to both sides

K^4+4q²=K²p²

divide both sides by K² and rearrange

p²=K²+4q²/K²

Thus, if we let a=K and b=2q/K, then p²=a²+b² as required by the problem. All that remains is to show that a and b are nonzero integers.

Lemma 1: K is an integer

Proof: All solutions of both quadratic equations are integers. K is the difference of two of these solutions - that is, K is the difference of two integers. Therefore, K is an integer.

Lemma 2: K is nonzero

Proof: K is the difference between a solution of the first equation and a solution of the second equation. Suppose K=0. Then these equations have a common solution - that is, for some x, x²+px+q=0=x²+px-q. This would imply that q=-q. This can only be true if q=0. But q≠0. Therefore, the equations do not have a common solution and K≠0.

Lemma 3: 2q/K is an integer

Proof: Earlier we showed that √(p²-4q)-√(p²+4q)=2K. We can perform some simple algebra as follows:

√(p²-4q)-√(p²+4q)=2K

multiply both sides by (√(p²-4q)+√(p²+4q)) and simplify

2K(√(p²-4q)+√(p²+4q)) = (p²-4q)-(p²+4q)

simplify further and rearrange:

-8q=2K(√(p²-4q)+√(p²+4q))

divide both sides by -4:

2q=-K(√(p²-4q)+√(p²+4q))/2

divide both sides by K:

2q/K=-(√(p²-4q)+√(p²+4q))/2

Note that 2K=√(p²-4q)-√(p²+4q) Adding 2√(p²+4q) to both sides of that identity gives 2K+2√(p²+4q) = √(p²-4q)+√(p²+4q). We can then substitute to get:

2q/K=-(2K+2√(p²+4q))/2

Simplifying at last yields:

2q/K=-K-√(p²+4q)

We already know K is an integer, so if √(p²+4q) is also an integer, then 2q/K is the difference of two integers and therefore an integer. Now we know that the solutions of the second equation are (-p+√(p²+4q))/2 and (-p-√(p²+4q))/2. The difference between them is √(p²+4q). However, since both of the solutions to the second equation are integers, their difference must be an integer. Therefore √(p²+4q) is an integer, and by extension, 2q/K is an integer. Q.E.D

Lemma 4: 2q/K is nonzero

We know 2q/k=-K-√(p²+4q). Suppose 2q/K=0. Then -K-√(p²+4q)=0 From this we would get:

-K-√(p²+4q)=0

add K to both sides and rearrange:

K=-√(p²+4q)

multiply both sides by 2

2K=-2√(p²+4q)

remembering 2K=√(p²-4q)-√(p²+4q), we substitute:

√(p²-4q)-√(p²+4q)=-2√(p²+4q)

add √(p²+4q) to both sides and rearrange:

-√(p²+4q)=√(p²-4q)

However we specified in the beginning that K was the difference between the two greater solutions of each equation - that is, the sign on the square roots of both discriminants is nonnegative. Therefore, -√(p²+4q)=√(p²-4q) would imply that √(p²+4q)=0=√(p²-4q). However, if √(p²+4q)=0=√(p²-4q), q=-q and q=0. But q≠0, therefore -√(p²+4q)≠√(p²-4q) and 2q/K≠0. Q.E.D

To sum up, we have shown that if we let a=K and b=2q/K, a and b will be nonzero integers such that a²+b²=p². Since K exists for all p, q; a and b integers satisfying a²+b²=p² (a≠0, b≠0) exist for all p, q satisfying the initial conditions.

Quod Erat Demonstrandum.

Edit: as a general rule, the first proof you come up with for something will not be the most efficient one. This has certainly proved true here. Lemma 4 can be proven in a much less circumlocutious manner by noting that if 2q/K=0, 2q=0 and q=0, but q≠0, thus neither is 2q/K. I'll leave the original proof of lemma 4 up, just so you can see how utterly silly some people are when trying to prove mathematical theorems.

Your question is troubled. If the equations are as you claim 'independent' then you mean X in one is not X in the other. Hence

X^2 + pX + q = 0

Y^2 + pY - q = 0

These two equations are independent, with equivalent coefficients. We can solve for p yielding:

p = -(X^2 + Y^2)/(X + Y) = Y - X

Are you asking if a and b are solutions to X and Y? Then clearly

p^2 = a^2 - 2ab + b^2

and not the relationship you specify if this is what was intended.

The situation as described seems contradictory: in particular, it's not possible for the two equations to hold unless q = 0 (to prove it, subtract the 2nd equation from the first), but we're also supposed to assume that q <> 0. Sure you typed it right?

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