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If the following quadratic equation $$qx^2+(p+q)x+bp=0$$ always has rational roots for any non-zero integers $p$ and $q$ what will be the value of $b$?

My book's solution says the value of $b$ will be $0$ or $1$.

If we consider the discriminant of the equation, $$D=(p+q)^2-4bqp = p^2+2q(1-2b)p+q^2$$ then $D$ should be a perfect square of a rational number for the equation having rational roots. so the value of b should be 0 0r 1 for given conditions (p and q are non-zero integers and the equation always has rational roots)

But I am not sure whether we can conclude the discriminant $D=p^2+2q(1-2b)p+q^2$ is a perfect square of a rational number if only if b is 0 or 1.

I can see if b is 0 or 1, then $D$ will be $(p+q)^2$ or $(p-q)^2$. So $D$ will be a perfect square of a rational number.

But I can't figure out the other case: If $D$ is a perfect square of a rational number for any non-zero integer p and q , then the value of b will be 0 or 1.

EDIT 1: b is rational number.

MS.Kim
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2 Answers2

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The key is that it must always have rational roots for any $p, q$.

Claim: Let $f(x) = x^2 + bx + c$, where $b$ and $c$ are rational numbers. If $f(x)$ satisfies the condition that for all rational numbers $r$, $f(r)$ is the square of a rational number, then $f(x)$ must be the square of a polynomial.

Proof: By completing the square, we simply need to consider polynomials of the form $x^2 + \frac{n}{m}$, where $m,n$ are integers and $\gcd(n,m) = 1$.

Consider $x = 0$ and $\frac{1}{m}$, which gives us $\frac{n}{m}$ and $\frac{1+mn}{m^2}$ are both squares of rational numbers. Multiplying both terms by $m^2$, we get that $mn$ and $mn+1$ are both integers that are squares of a rational number, hence they are perfect squares, so $mn=0$.

Since $m\neq 0$, hence $n=0$ and thus $f(x) = x^2$ and we are done. $_\square$

Corollary: $b=0$ or $1$.

Proof: Consider the polynomial $f(x) = x^2 + 2(1-2b) x + 1$. From the conditions (see OP's work), for rational numbers, this always evaluates to a rational square. Hence, we must have $ f(x) = (cx+d)^2$.

Comparing coefficients, we see that $ c ^2 = 1, d^2 = 1$, which implies that $f(x) = (x+1)^2 $ or $(x-1)^2$, which hence correspond to $b=0$ and $b=1$ respectively. $_\square$

Note: Franklin shows that the claim is true for all polynomials, though it uses more machinery.

Calvin Lin
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Here's an approach that doesn't appeal to knowing the existence of a high-power theorem (even if it can be proved via elementary means).

Idea: Makes the perfect square an integer. Bound it between perfect squares to narrow down the value. (We couldn't have done this with perfect square rationals which are dense on the real line.)

Show that

  1. $b$ is rational.
  2. If $b = \frac{m}{n}$, show that by considering $p \in n\mathbb{Z}$, it is sufficient to prove it when $b$ is an integer.
    • This step isn't necessary here as it can be combined with the next. However, such simplifications are useful to think about, hence I'm calling it out explicitly.
  3. As OP said, we want the discriminant $D=p^2+2q(1-2b)p+q^2$ to be a perfect square.
    • Since $D$ is now an integer, hence it is a perfect square integer.
    • $D = (p+(1-2b)q)^2 - 4b(1-b)q^2$
  4. Take $ q = 1$ (We just need to fix a non-zero $q$.) For each $p$, let $ D_p = [p+(1-2b) - c_p]^2$, where $c_p$ is an integer.
    • For $|p| >> |(1-2b)|$, $D - (p+(1-2b))^2 = 4b(1-b)$ is a fixed quantity, hence eventually has absolute value smaller than $2|p+1-2b|$, IE $$(p+(1-2b) - 1)^2 < (p+(1-2b)q)^2 - 4b(1-b)q^2 < (p+(1-2b) +1)^2.$$
    • This show that $c_p = 0$ for all large enough values.
  5. Hence $4b(1-b) = 0$, or that $ b = 0, 1$.
Calvin Lin
  • 77,541