One Symbol Changes Everything

Relevant wiki: Quadratic Diophantine Equations - Solve by Factoring

Another approach is as follows. Let $a,b$ be the roots of $x^{2} - px + q = 0$ and $c,d$ be the roots of $x^{2} - qx + p = 0$ . Then by Vieta's we have that

$a + b = p, ab = q, c + d = q, cd = p \Longrightarrow a + b = cd, c + d = ab$

$\Longrightarrow a + b + c + d = ab + cd \Longrightarrow ab - a - b + cd - c - d = 0$

$\Longrightarrow (a - 1)(b - 1) + (c - 1)(d - 1) = 2$ .

Now as both $a + b$ and $ab$ must be positive we must have both $a,b$ as positive integers. Likewise $c,d$ . Thus without loss of generality we can have either (i) $(a - 1)(b - 1) = (c - 1)(d - 1) = 1$ or (ii) $(a - 1)(b - 1) = 0$ and $(c - 1)(d - 1) = 2$ . Since the roots must be distinct we can rule out case (i), leaving us with case (ii) as the only viable option. Again without loss of generality we then set $a = 1, c = 3, d = 2$ , which then gives us $b = c + d = 5$ . Finally we find that $p + q = a + b + c + d = \boxed{11}$ .

Given equation are $x^2 -px + q =0$ (let this be equal to $f(x)=0$ ) and $x^2 -qx + p =0$ (let this be equal to $g(x)=0$ )

Now there are three possibilities ,

$p>q$

$p=q$ and

$p<q.$

$Case (I) : p > q$

$f(0)= q$

$f(1)= 1 - (p-q)$ Now in this case , if $p-q>1$ will mean that f(1) is negative. So since f(0) is positive and f(1) is negative will mean that there is a root between the number 0 and 1 . But wait we needed that the root must be and integer. So the only possibility is that $p-q = 1$ (which is the least difference possible since they are natural numbers) . so one root comes out to be 1 as f(1) will come out to be zero and by Vieta's product of roots formula we get other root as q which is also an integer (as q is natural number ). But we also needed the root of g(x)=0 to be integer.
So now let's check the discriminant of g(x)=0.

$D= q^2 - 4 p$ Substituting p=q+1 we get $D= q^2-4(q+1).$ Now for the root to be an integer the discriminant must be a perfect square. Therefore , let $q^2- 4p -4 = m^2$

$(q-2)^2 - m^2 = 8.$ Completing the square method is used to reach above step. Now since the difference between two square is 8. The only possibility comes out to be $3^2 - 1^2.$ ( Bonus :Can anybody prove it why is this the only possibility. I am also trying but not able to do it. One thing can be used that any square of integers is / are of form 4k or 4k+1). So q-2=3 => q= 5 and so p= 6. Okay now it comes the time for you to work. The case (iii) will give the same answer answer as the case (I) since both are kinda symmetric just the value will get reverse. So you got to prove that in case (ii) we won't get any solution .

Maths is all about learning and solving and then repeating it over and over again.