To prove that a given quadratic has integral roots

[brotherbobby]

Homework Statement

The equation $x^2+px+q = 0$, where $p \in \mathbb{Z}, q \in \mathbb{Z}\;$has rational roots. Prove that these roots are integers.

Relevant Equations

1. If $\alpha,\beta$ are the roots of the quadratic equation $x^2+px+q = 0$, then the sum of the roots $\alpha+\beta = -p$ and the product of roots $\alpha\beta = q$.

2. The roots of the quadratic equation $x^2+px+q = 0$ are rational if the discriminant $p^2-4q$ is a perfect square of an integer, i.e. $p^2-4q = r^2$ where $r \in \mathbb{Z}$.

Given : The quadratic equation $x^2+px+q = 0$ with coefficients $p,q \in \mathbb{Z}$, that is positive or negative integers. Also the roots of the equation $\alpha, \beta \in \mathbb{Q}$, that is they are rational numbers. To prove that $\boxed{\alpha,\beta \in \mathbb{Z}}$, i.e. the roots are integers too, like p and q.

Attempt : From properties of roots, $\alpha+\beta = -p$ and $\alpha\beta = q$. Also since the roots are rational, $p^2-4q = r^2$ where $r \in \mathbb{Z}$.

Thus we have $(\alpha+\beta)^2 - 4\alpha\beta = (\alpha-\beta)^2 = p^2-4q = r^2 \Rightarrow \alpha - \beta = \pm r$.

Thus $\alpha+\beta = -p$ and $\alpha - \beta = \pm r$.

This leads to $\alpha = \frac{-p+r}{2}$ and $\beta = -\frac{p+r}{2}$. (no surprise).

Clearly this does not prove that $\alpha,\beta$ are integers, unless I can show that both $p,r$ are multiples of 2.

I don't know how to proceed from here.

 

[Delta2]

Hint: Take cases
1)p is even,
2)p is odd

and examine what happens to $p^2-4q=r^2$ having in mind that 4q is even.
The following might be useful to you:
1) The square of an odd is also an odd number, the square of an even is an even number
2)Subtracting an even from an odd gives odd
3)Subtracting an even from an even gives even

 

[Delta2]

To hint a bit more your goal will be to prove that
1) if p is odd then r is odd too
2) if p is even then r is even too

In either of the above cases the roots will be integers (why?).

 

[PeroK]

Why not let $\frac m n$ be a solution, plug that in and see what happens? All you have to do is show that $n = 1$.

 

[Delta2]

Well @anuttarasammyak had posted the whole problem solution using Perok's suggestion but he deleted his message , Perok's method works indeed and is faster no need to consider cases there.

 

[brotherbobby]

Hint: Take cases
1)p is even,
2)p is odd

and examine what happens to $p^2-4q=r^2$ having in mind that 4q is even.
The following might be useful to you:
1) The square of an odd is also an odd number, the square of an even is an even number
2)Subtracting an even from an odd gives odd
3)Subtracting an even from an even gives even

Thank you, the solution seems obvious, but let me put it nonetheless for completeness' sake.

We had arrived at the following equation for the roots to be rational (the discriminant be a perfect square) : $p^2-4q = r^2$. From here (see my post#1), we had the two (rational) roots as $\alpha = \frac{-p+r}{2}, \beta = \frac{-p-r}{2}$.

Of course $4q$ is always even. But how about $p$?

(1) $\boldsymbol {p}$ is even : If $p$ is even, then $r^2 = p^2-4q \rightarrow \; \text{even}\; \Rightarrow r\rightarrow\; \text{even}$ which make both the roots $\alpha,\beta$ even since the sum and difference of even numbers are even.

(2) $\boldsymbol p$ is odd:
If $p$ is odd, then $r^2 = p^2-4q \rightarrow \; \text{odd}\; \Rightarrow r\rightarrow\; \text{odd}$ which make both the roots $\alpha,\beta$ even since the sum and difference of odd numbers are even.

Thank you.

 

[brotherbobby]

Well @anuttarasammyak had posted the whole problem solution using Perok's suggestion but he deleted his message , Perok's method works indeed and is faster no need to consider cases there.

I couldn't work out the problem this way. Any hints?

 

[PeroK]

I couldn't work out the problem this way. Any hints?

$$\frac{m^2}{n^2} + p\frac m n + q = 0 \ \Rightarrow \ m^2 + pmn + qn^2 = 0$$
Now, assume $\frac m n$ is in its lowest form.

 

[brotherbobby]

$$\frac{m^2}{n^2} + p\frac m n + q = 0 \ \Rightarrow \ m^2 + pmn + qn^2 = 0$$
Now, assume $\frac m n$ is in its lowest form.

From the equation $m^2+pmn+qn^2=0$, we know that $m \in \mathbb{Z}$. Hence using $m$ as the variable (like $x$) and solving for it using the quadratic formula : $m = \frac{-pn\pm \sqrt{p^2n^2-4qn^2}}{2}\Rightarrow m = \frac{-pn\pm nr}{2}$ where $r = \sqrt{p^2-4q} \in \mathbb{Z}$. Hence we have $m = \frac{-n(p \mp r)}{2}$.

I don't see why $n = 1$?

 

[PeroK]

From the equation $m^2+pmn+qn^2=0$, we know that $m \in \mathbb{Z}$. Hence using $m$ as the variable (like $x$) and solving for it using the quadratic formula : $m = \frac{-pn\pm \sqrt{p^2n^2-4qn^2}}{2}\Rightarrow m = \frac{-pn\pm nr}{2}$ where $r = \sqrt{p^2-4q} \in \mathbb{Z}$. Hence we have $m = \frac{-n(p \mp r)}{2}$.

I don't see why $n = 1$?

$$ m^2 + pmn + qn^2 = 0 \ \Rightarrow \ m^2 = -n(pm + qn)$$
Does that give you any ideas?

 

[Charles Link]

a hint would be to prime factor both sides of the last expression. What form must $n$ and the integer factor next to it have compared to $m^2$? Would it be possible for $n$ to not have factors in $m$?

 

[PeroK]

... if we let $k = -(pm + qn)$, then we have:$$m^2 = nk \ \ \text{and} \ \ \frac{m^2}{n} = k$$ Yet, $m, n$ are supposed to have no common factors. Is this possible?

 

[Delta2]

It is a well known result from number theory that if a number $x$ divides $y^k$ (for any natural number k) then $x$ and $y$ are not coprimes, the proof of this result relies on the fundamental theorem of number theory as @Charles Link suggests.

 

[brotherbobby]

... if we let $k = -(pm + qn)$, then we have:$$m^2 = nk \ \ \text{and} \ \ \frac{m^2}{n} = k$$ Yet, $m, n$ are supposed to have no common factors. Is this possible?

Hi all, the creator of this thread here and sorry for coming in late to the discussion.

I get that $m/n$ has no common factor, that is $\frac{m}{n} = k \not\in \mathbb{Z}$.

However, the argument so far goes like this :

If $x = \frac{m}{n}$ then $x^2+px+q = 0 \Rightarrow \frac{m^2}{n^2}+p\frac{m}{n}+q = 0\Rightarrow m^2+pmn+n^2q = 0\Rightarrow m^2 = -n(pm+nq)\Rightarrow m^2 = nk'$ where $k' = -(pm+nq)$.

From above, we have $m \frac{m}{n} = k'$ which might be an integer, i.e. $k' \in \mathbb{Z}$.

We note from above that while $\frac{m}{n} = k \not\in \mathbb{Z}$, $k' = mk$ which may be an integer.

I don't see how we can use the above argument to show that $n = 1$.

(If $m,n$ have no common factor $k (\in \mathbb{Z})$, can we use that to say $m^2,n$ also have no common factor?)

 

[PeroK]

(If $m,n$ have no common factor $k (\in \mathbb{Z})$, can we use that to say $m^2,n$ also have no common factor?)

Can you find an example where $m, n$ have no common factor, but $m^2, n$ do? If not, why not? Hint: think about prime factorisation.

 

[brotherbobby]

Yes I can find no example where $m,n$ have no common factors and yet $m^2,n$ do. However, this does not constitute a proof.

Can you explain prime factorisation to me? I am aware that two numbers $m,n$ are said to be mutually prime if they have no integrer that divides them. Of course this does not mean $m,n$ are prime numbers themselves. For example, 12 and 7 are mutually prime to one another. A better example would be 14 and 9. Still, I am not clear on this point. If two numbers $m,n$ are not mutually prime, does it mean that their common factor has to be a prime number?

Thank you for your time.

 

[PeroK]

Yes I can find no example where $m,n$ have no common factors and yet $m^2,n$ do. However, this does not constitute a proof.

Can you explain prime factorisation to me? I am aware that two numbers $m,n$ are said to be mutually prime if they have no integrer that divides them. Of course this does not mean $m,n$ are prime numbers themselves. For example, 12 and 7 are mutually prime to one another. A better example would be 14 and 9. Still, I am not clear on this point. If two numbers $m,n$ are not mutually prime, does it mean that their common factor has to be a prime number?

Thank you for your time.

The key point is that every integer, except $0$ and $\pm 1$, has a prime factor. Therefore: if two numbers have a common factor, they must have a prime common factor. E.g. if their common factor is $30$, then they have $2, 3$ and $5$ as prime common factors.

 

[PeroK]

PS The question in the OP seems to me to be clearly in the field of number theory. If you haven't studied the basics, like prime factorisation, then that sort of question is going to be very awkward.

PPS that's why unique prime factorisation is called the fundamental theorem of arithmetic!

 

[anuttarasammyak]

In other words
$$gcd(m,n)=1$$
In factorization m and n do not have common factors other than 1. All the rational numbers are written in form of $\frac{m}{n}$ where $gcd(m,n)=1$. They are called coprime. Obviously we get

$$gcd(m,n)=1 \Leftrightarrow gcd(m^k,n)=1$$

For an example
$$gcd(2,3)=1=gcd(4,3)=gcd(8,3)=gcd(8,9)=gcd(8,27)$$

 

[mathwonk]

use these ideas to prove the "rational roots theorem" from high school algebra: if x = n/m is a rational root in lowest terms, of the equation aX^n + ...+c = 0, where all coefficients a,..,c are integers, then m is a factor of a and n is a factor of c. Consequently, every rational root of a "monic" equation X^n+bX^(n-1)+...+c =0, with all coefficients integers, is itself an integer.

 

[Delta2]

The key to notice is that the prime factors of $m$ and $m^k$ ($k$ natural number, in this case $k=2$) are the same, they are just raised to the $k$ power. So if $n$ divides $m^k$ , it means $n,m^k$ have a common prime factor and because the prime factors of $m$ and $m^k$ are the same, we conclude that $n$ and $m$ hace a common prime factor.