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I am having trouble with this tricky question on number theory and quadratic equations.

If $a$ and $c$ are odd prime numbers and $ax^2+bx+c=0$ has rational roots, where $b\in \Bbb Z$, then prove that one root of the equation will be independent of $a$ and $b$.

So I thought of trying the following way:

If it has rational roots (say $u$ and $m$) then $um = ac$ and $u + m = b$. As $a$ and $c$ are prime, $a + c = b$, and plugging this into the quadratic formula, we get one root will always be $-1$.

I am not sure if this proof is correct or if my final value is correct. Could someone please give a detailed proof?

P.S.: I couldn’t find any examples regarding my question so if you do, please include them too.

Bill Dubuque
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Aditya_math
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5 Answers5

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Since $ \ a \ $ and $ \ c \ $ are odd prime numbers, the Rational Zeroes Theorem proposes the candidates $$ \ x \ = \ \pm \frac{c}{a} \ , \ \pm \frac{c}{1} \ , \ \pm \frac{1}{a} \ , \ \pm \frac{1}{1} \ \ . $$

The factorization of the quadratic polynomial is $ \ ax^2 \ + \ bx \ + \ c \ = \ a·(x - r)·(x - s) \ \ , $ so the four factorizations that will produce the correct leading and constant terms are $$ a·\left(x - \frac{c}{a} \right)·\left(x - 1 \right) \ \ , \ \ a·\left(x + \frac{c}{a} \right)·\left(x + 1 \right) \ \ , \ \ a·\left(x - \frac{1}{a} \right)·\left(x - c \right) \ \ , $$ $$ a·\left(x + \frac{1}{a} \right)·\left(x + c \right) \ \ . $$

The "middle coefficient" $ \ b \ = \ -a·(r + s) \ $ in these products of factors is one of $$ -a · \left(\pm \frac{c}{a} \ \pm \ 1 \right) \ \ = \ \ \pm \ (c + a) \ \ \ \text{or} \ \ \ -a · \left(\pm \frac{1}{a} \ \pm \ c \right) \ \ = \ \ \pm \ (1 + ac) \ \ , $$ any of which is an integer.

Hence, one of the rational roots of the quadratic equation is $ \ \pm 1 \ $ or $ \ \pm c \ \ , $ which are independent of $ \ a \ $ and $ \ b \ \ . $

  • No need to factor $f(x)!::!$ after RZT, by Vieta the roots have product $,\large\frac{c}a,$ so the two roots are $,\large \frac{c}a, \color{#c00}{\frac{1}1}$ or $,\large\color{#c00}{\frac{c}1},\frac{1}a$ (or their negatives), so one of the roots is an integer $,\color{#c00}{\pm1}$ or $\color{#c00}{\pm c},,$ so is "independent" of $,a,b.\ \ $ – Bill Dubuque Apr 03 '25 at 05:07
  • Simpler: if both roots were non-$\color{#c00}{\style{font-family:inherit;}{\text{integers}}}$ then by RZT they are $,i/a,,j/a,,$ prime $,a\nmid i,j,$ so $,a\nmid ij,,$ so the root product has least denom $a^2,,$ contra product = $c/a$ by Vieta. So at least one root is an $\color{#c00}{\style{font-family:inherit;}{\text{integer}}}$ $,n,,$ thus RZT $\Rightarrow ,n\mid c,$ prime, so $,n=\color{#c00}{\pm1}$ or $\color{#c00}{\pm c},$ is "independent" of $,a,b.,$ I added an answer which explains from a more general perspective why such proofs work. $\ \ $ – Bill Dubuque Apr 03 '25 at 05:08
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I think you are almost correct (assuming you mean by $I$ the set of integers). The only thing you miss is that if $a,c$ are prime, then you have $a+c=\pm b$ (which follows from the fact that $b^2-4ac$ must be the square of a rational number), so the possible values are $\pm 1$. You can get as many examples as you want just from the formula $a+c=\pm b$ you have (e.g. $(a,b,c)=(3,\pm 8,5)$).

P.s. Your formulas u * m = a * c and u + m = b look wrong, but if you have $a+c=\pm b$, your quadratic equation is just $ax^2 \pm (a+c) x + c =0$, which you can just solve and is essentially the same as using the formula you want to apply.

P. Usada
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  • $b^2-4ac$ must be the square of an integer, $b^2-4ac=d^2$, say, so $4ac=b^2-d^2=(b+d)(b-d)$, so $2a$ and $2c$ are $\pm(b\pm d)$ with some appropriate choice of signs. From that, you can get $a+c=\pm b$. – Gerry Myerson Jun 14 '20 at 09:59
  • I think a priori you only know that it is a square of a rational number. The fact that your $d$ is an integer follows from the supposition that $b$ is an integer. The rest of arguments is the same as what I thought. – P. Usada Jun 14 '20 at 10:10
  • You're given that $a,b,c$ are all integers, whence $b^2-4ac$ is an integer, and it's a standard result that if an integer is the square of a rational, then it's actually the square of an integer. – Gerry Myerson Jun 14 '20 at 10:17
  • I do not know why you are so aggressive, but I am agreeing with everything you are saying. – P. Usada Jun 14 '20 at 10:22
  • My apologies, no aggression was intended. I am glad we are in agreement on the mathematics. – Gerry Myerson Jun 14 '20 at 10:26
  • @PekoraUsada thank you so much, so is the answer supposed to be -1, because i tried it this way first and I was getting -1 as the answer so I tried another method and got the same answer so I thought both were wrong, thank you so much though – Aditya_math Jun 14 '20 at 11:30
  • @Gerry In your first comment, how did you infer $:!4ac = (b+d)(b-d),\Rightarrow$ $, a,c\mid b\pm d,,$ i.e how do you rule out the case $,ac,$ divides one of $,b\pm d,,$ i.e. both primes occur in one factor? $\ \ $ – Bill Dubuque Apr 03 '25 at 21:58
  • @P.Usada Same question for you - cf. my prior comment. $\ \ $ – Bill Dubuque Apr 04 '25 at 07:14
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More generally, the Lemma below implies some rational root of OP's $f(x)\,$ is an integer $\,n,\,$ so the Rational Root Test $\,\Rightarrow\,n\mid f(0)\!=\!c\,$ prime, so $\, n = \pm1,\:\!\pm c,\,$ which is "independent" of $\,a,b$.

Lemma $ $ An integer coef polynomial $f(x)$ has an integer root if it has $\rm\color{#c00}{more}$ rational roots than its leading coef has primes in its prime factorization (including multiplicity in both counts).

Proof $\ $ Let $\,{\large\frac{e_i}{d_i}}\in\Bbb Q,\,\ d_i \! > \!0\,$ be the reduced roots. By the Factor Theorem and Gauss's Lemma

$$f(x) \,=\, (d_1 x - e_1)\cdots (d_k x - e_k)\, g(x),\,\ {\rm for}\,\ g(x)\in \Bbb Z[x]\qquad$$

Comparing lead coefs shows the lead coef of $f\:\!$ is divisible by $\:\!\prod_i d_i$. If all $\,d_i\neq 1$ then each $d_i$ contributes at least one prime factor to the lead coef, contra $\rm\color{#c00}{hypothesis}$. Hence some $\,d_i = 1,\,$ so $\,x = e_i\,$ is an integer root of $\,f.\,$ $\,\bf\small QED$

Bill Dubuque
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  • The Lemma can be viewed as generalization of the Rational Root Test from one to many rational roots, since it implies that the product of the reduced denominators divides the lead coef. $\ \ $ – Bill Dubuque Apr 04 '25 at 07:03
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Let $y=ax.$ Then, $$y^2+by+ac=0.\tag1$$ Let $y_1=\frac pq$, $\gcd(p,q)=1$ one of the rational roots of $(1)$. By Vieta's formulae, $-p(p+bq)=acq^2$. So, $q=1.$ The possible solutions of $(1)$ are

  1. $(\pm1,\pm ac)$
  2. $(\pm a,\pm c)$
  3. $(\pm c,\pm a)$
  4. $(\pm ac, \pm 1)$

Then the claim in OP follows.

Bob Dobbs
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  • Please explain more explicitly in the answer how "the claim follows". It will likely be unclear to many readers. – Bill Dubuque Apr 01 '25 at 20:09
  • @BillDubuque I edited about, your request. – Bob Dobbs Apr 01 '25 at 20:42
  • You still haven't explicitly explained how it follows for a root $x$ of $f(x).,$ There is no need to scale to integer roots using that AC method since RRT = Rational Root Test already yields a simple proof - cf. user882145's answer (you are essentially scaling that proof by $a).,$ RRT and AC are equivalent in this context, so we can always replace use of RRT by AC, but usually doing so is more roundabout (a bit less so here since lead & constant coefs are primes). $\ \ $ – Bill Dubuque Apr 01 '25 at 21:30
  • Simpler: by ac method: scaling the roots $,r,r'$ by $,a,$ yields integer roots with product $,(ar)(ar')=\color{#0a0}ac,,$ so $,\color{#0a0}a,$ prime $\Rightarrow \color{#0a0}a\mid ar,$ or $,\color{#0a0}a\mid ar',,$ i.e. $,r\in\Bbb Z,$ or $,r'\in\Bbb Z.,$ wlog $,r\in\Bbb Z,,$ so $,r,$ divides prime $,c = -r(ar+b),$ so $,r = \pm1,\pm c,$ is "independent" of $,a,b.,$ See also my comments on user882145's answer. $\ \ $ – Bill Dubuque Apr 03 '25 at 08:56
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Let $\alpha$ and $\beta$ be roots of the quadratic equation
$\therefore$ $\alpha$ and $\beta$ $\in \Bbb{Q}$ (where $\Bbb{Q}$ represents set of rational numbers)
$\therefore$ $b^2-4ac$ is a perfect square (as if it is not a perfect square, roots will not be rational)

Let $b^2-4ac = \lambda^2$
$\therefore$ $\sqrt{b^2-4ac}=\lambda$ where $\lambda \in \mathbb{W}$ (as output of square root function is always non-negative and also $-4ac$ should be a negative integer as $a$ and $c$ are odd primes, therefore $b^2$ should be a natural number as $b$ is an integer, ($b$ cannot be zero otherwise $b^2-4ac$ will be negative) and therefore, $b^2-4ac$ should be a whole number)

$\therefore b^2-\lambda^2 = 4ac$
$\therefore (b+\lambda)(b-\lambda) = 4ac$

$(b+\lambda)$ and $(b-\lambda)$ both can be odd or both can be even
$\because$ Product of $(b+\lambda)$ and $(b-\lambda)$ is even $\therefore (b+\lambda)$ and $(b-\lambda)$ are even

$\therefore$ Case 1 $\implies$ $b+\lambda=2a$ , $b-\lambda=2c$ $\implies$ $b=a+c$ , $\lambda=a-c$
Case 2 $\implies$ $b+\lambda=2c$ , $b-\lambda=2a$ $\implies$ $b=a+c$ , $\lambda=-a+c$
Case 3 $\implies$ $b+\lambda=-2a$ , $b-\lambda=-2c$ $\implies$ $b=-a-c$ , $\lambda=-a+c$
Case 4 $\implies$ $b+\lambda=-2c$ , $b-\lambda=-2a$ $\implies$ $b=-a-c$ , $\lambda=a-c$
Case 5 $\implies$ $b+\lambda=2$ , $b-\lambda=2ac$ $\implies$ $b=1+ac$ , $\lambda=1-ac$
Case 6 $\implies$ $b+\lambda=-2ac$ , $b-\lambda=-2$ $\implies$ $b=-1-ac$ , $\lambda=1-ac$
Case 7 $\implies$ $b+\lambda=-2$ , $b-\lambda=-2ac$ $\implies$ $b=-1-ac$ , $\lambda=ac-1$
Case 8 $\implies$ $b+\lambda=2ac$ , $b-\lambda=2$ $\implies$ $b=1+ac$ , $\lambda=ac-1$

Cases 5 and 6 are eliminated as lambda is not a natural number for these cases $\because$ $b+\lambda < b-\lambda$ and/or $\lambda = 1 - ac ≤ -8$ because $ac$ has minimum value of 9 as minimum value of $a$ and $c$ is 3

By Quadratic Formula,
$ x = {-b \pm \sqrt{b^2-4ac} \over 2a} $

$\therefore x = {-b \pm \lambda \over 2a} $

$\therefore x = {-b + \lambda \over 2a}$ or ${-b - \lambda \over 2a} $

$\therefore x = {-(b - \lambda) \over 2a}$ or ${-(b + \lambda) \over 2a} $

$\therefore$ In case 1 and case 2 ; $x$ = $-c \over a$ or -1
$\therefore$ In case 3 and case 4 ;$x$ = $c \over a$ or 1
$\therefore$ In case 7 ; $x$ = $c$ or $1 \over a$
$\therefore$ In case 8 ; $x$ = $-c$ or $-1 \over a$

Therefore, if we see cases 1-4, case 7 and case 8, we can conclude that one root will be independent of a and b.

user147259
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  • Simpler we can eliminate those $8$ cases: you have $,ac = \color{#0a0}{(-b+\lambda)/2} \cdot \color{#90f}{(-b-\lambda)/2},,$ which is a product of integers by $!\bmod 2!:\ 0\equiv 4ac= b^2-\lambda^2\equiv b-\lambda.,$ But by the quadratic formula $,(-b\pm\lambda)/(2a),$ are the roots $,r_i,$ of $,f(x),$ hence our first equation becomes $,ac = \color{#0a0}{(ar_1)}\color{#90f}{(ar_2)},,$ from which it is very easy to finish. $\ \ $ – Bill Dubuque Apr 04 '25 at 23:50