Find the integer pair (a,b) that satisfies this

Question

I need some help finding a pair of integers $(a,b)$ that satisfies this constraint: $x^2 - ax - b = 0$ has integer solutions and $x^2 +ax+b =0$ has integer solutions. The pair $(5,6)$ satisfies this because $x^2 - 5x - 6 = 0$ has integer solutions i.e. x = {-1,6} and $x^2 + 5x + 6 = 0$ has integer solutions i.e. x = {-3,-2}.

If there is no general form can you help me to at least find another pair of integers that satisfies this constraint other than $(5,6)$? Thanks in advance.

EDIT: the roots of the quadratic equations $x^2 - ax - b = 0$ hand $x^2 +ax+b =0$ can be rational also. The need not be constrained to integers. Just make sure they aren't irrational. The pair (a,b) however must be integers.

Answer 1

Let the discriminants of the two equations be $a^2+4b=k^2$ and $a^2-4b=l^2$. Note that

• $a,k,l$ have the same parity, so the discriminants being perfect squares is enough to ensure integer solutions
• the difference of two squares of the same parity is always a multiple of $4$, so we need not worry about $b$ being non-integral – it suffices to get $a,k,l$

Now $l^2,a^2,k^2$ are perfect squares in arithmetic progression, so $l^2+k^2=2a^2$ or $(l/a)^2+(k/a)^2=2$. All rational solutions for $l/a$ and $k/a$ may be parametrised as follows: $$\frac la=\frac{m^2-2m-1}{m^2+1}\qquad\frac ka=\frac{m^2+2m-1}{m^2+1}$$ where $m\in\mathbb Q\cup\{\infty\}$.

Given a rational solution for $\frac la$ and $\frac ka$, write the fractions with a common denominator, from which $l,a,k$ are immediately determined up to sign (their signs must all agree). From there $b$ may be easily worked out.

Answer 2

The product of the roots of one equation must be the negative of the product of the other pair of roots. Also, the sum of the roots of one equation must be the negative of the sum of the other pair of roots.

A formula for this is to let the pairs of roots be

$L^2-3LM+2M^2,LM$ and $M(L-2M),L(M-L)$

for integers $L$ and $M$.

Answer 3

If we call $ \ f(x) \ = \ x^2 + ax + b \ \ , \ $ then $ \ x^2 - ax - b \ = \ f(-x) - 2b \ = \ g(x) \ \ ; \ $ the parabola representing this second function is a reflection about the $ \ y-$axis of the parabola for the first function, shifted "downward" by $ \ 2b \ $ units. As a consequence, we will only consider solutions with $ \ a \ > \ 0 \ \ , \ $ as the corresponding solution with $ \ a \ < \ 0 \ $ is not distinct.

Plainly, there are "trivial" solutions with $ \ b \ = \ 0 \ \ $ being an integer, so that the pair of functions have the solution sets $ \ \{ 0 \ , \ -a \} \ $ and $ \ \{ 0 \ , \ +a \} \ \ . \ $ Otherwise, there is a constraint imposed on $ \ b \ $ in order that both polynomials have real zeroes. The "vertex form" of the two function expressions is $ \ f(x) \ = \ \left(x + \frac{a}{2} \right)^2 \ + \ \left(b - \frac{a^2}{4} \right) \ \ $ and $ \ g(x) \ = \ \left(x - \frac{a}{2} \right)^2 \ + \ \left(-b - \frac{a^2}{4} \right) \ \ , \ $ so we require $ \ b \ \le \ \frac{a^2}{4} \ \ ; \ $ the non-trivial solutions thus occur for $ \ a \ \ge \ 2 \ \ . \ $

The next most simple case to examine is for $ \ b \ = \ \frac{a^2}{4} \ \ , \ $ for which $ \ f(x) \ $ has the solutions $ \ \{ -\frac{a}{2} \ , \ -\frac{a}{2} \} \ \ . \ $ We then have $ \ g(x) \ = \ x^2 - ax - \frac{a^2}{4} \ \ , \ $ the discriminant being $ \ \Delta \ = \ 2·a^2 \ \ . \ $ But $ \ a \ $ is required to be an integer, so $ \ g(x) \ $ cannot have rational zeroes for this case; hence, we must have the strict inequality $ \ b \ < \ \frac{a^2}{4} \ \ . $

The discriminants for $ \ f(x) \ $ and $ \ g(x) \ $ are in general $ \ \Delta_f \ = \ a^2 - 4b \ $ and $ \ \Delta_g \ = \ a^2 + 4b \ \ , \ $ both of which must be perfect squares if the zeroes of both functions are to be rational. We will wish to find integer triples $ \ b \ , \ \sqrt{a^2 - 4b} \ , \ a \ \ $ and $ \ b \ , \ a \ , \ \sqrt{a^2 + 4b} \ \ . \ $ For your example, $$ \ a \ = \ 5 \ \ , \ \ b \ = \ 6 \ \ \Rightarrow \ \ \sqrt{5^2 - 4·6} \ = \ 1 \ \ , \ \ \sqrt{5^2 + 4·6} \ = \ 7 \ \ $$ (producing the solution sets $ \ \{ \ -\frac52 \ \pm \ \frac12 \ \} $ and $ \ \{ \ +\frac52 \ \pm \ \frac72 \ \} \ ) \ . $

This proves to be an extremely restrictive condition. In order that these square-roots be integers, we need two perfect squares having a difference of a multiple of $ \ 8 \ \ . \ $ Since "consecutive squares" differ by odd integers, the required squares must be an even number of positions apart; again for your example, we have $ \ 1 \ \ (4 \ \ 9 \ \ 16) \ \ 25 \ \ (36) \ \ 49 \ \ $ (we also observe that this is the "smallest" set of squares that meet the requirements). This set is the prototype of a "family" of solutions: from a difference between squares of $ \ 24 \ \ , \ $ we find another set with a difference of $ \ 96 \ \ , \ $ $ 4 \ , \ 100 \ , \ 196 \ \ , \ $ corresponding to $$ \ a \ = \ 10 \ \ , \ \ b \ = \ 24 \ \ \Rightarrow \ \ \sqrt{5^2 - 4·6} \ = \ 2 \ \ , \ \ \sqrt{5^2 + 4·6} \ = \ 14 \ \ $$ and the polynomials $ \ x^2 + 10x + 24 \ = \ (x + 4)·(x + 6) \ \ , \ \ x^2 - 10x - 24 \ = \ (x + 2)·(x - 12) \ \ . \ $ We thus have the family of polynomial pairs $ \ f(x) \ = \ x^2 + 5mx + 6m^2 \ = \ (x + 2m)·(x + 3m) \ \ , $ $ g(x) \ = \ x^2 - 5mx - 6m^2 \ = \ (x + m)·(x - 6m)\ = \ f(-x) - 12m^2 \ \ , \ m \ $ being a positive integer (since we chose $ \ a \ > \ 0 \ ) \ \ . $

I have not conducted an extensive search for other possible families, so I am not sure at this point whether this family is unique (neglecting the "trivial family"). My suspicion is that it may well be due to the relation between the functions, $ \ g(x) \ = \ f(-x) - 2b \ . $