Problems on Integer Roots

In one sentence

The quadratic formula contains a square root:

x = \frac{- b \pm Δ}{2 a}

When the coefficients are all integers and you want $x$ to be an integer, the square root has to vanish first: $Δ$ must be a perfect square. That is the first hurdle in every integer-root problem. Once you are past it, the numerator still has to be divisible by $2 a$ .

Predict first

Both roots of $x^{2} - 6 x + c = 0$ are positive integers. How many possible values of $c$ are there?

2

3

5

6

Choose an answer first; then you can see the result.

The discriminant route: when does x² − mx + m + 2 = 0 have integer roots?

Step 1 / 5

Try one first
$m = 1$ : $Δ = 1 - 12 = - 11 < 0$ , so there are not even any real roots.

$Δ = m^{2} - 4 (m + 2) = m^{2} - 4 m - 8$
Real roots are not enough
Let Δ = k² and complete the square
Split 12 into a product of two integers
Check the roots

Stop 1 · Vieta's formulas: no discriminant needed

Same problem, different road.

One problem, several solutions

$m$ is an integer and both roots of $x^{2} - m x + m + 2 = 0$ are integers. Find $m$ .

The discriminant route

$Δ = (m - 2)^{2} - 12 = k^{2}$ , so $(m - 2 - k) (m - 2 + k) = 12$ . The two factors have the same parity, so they can only be $2 \times 6$ or $(- 6) \times (- 2)$ , giving $m = 6$ or $m = - 2$ . Checking, the roots are integers in both cases.

CostWorks in general; needs “difference of squares + splitting into factors”, and you must check the roots.

The Vieta route

Call the roots $x_{1}$ and $x_{2}$ : $x_{1} + x_{2} = m$ and $x_{1} x_{2} = m + 2$ . Subtract to eliminate $m$ :

x_{1} x_{2} - x_{1} - x_{2} = 2 \Rightarrow (x_{1} - 1) (x_{2} - 1) = 3.

$3 = 1 \times 3 = (- 1) \times (- 3)$ , so the roots are $2, 4$ or $0, - 2$ , giving $m = 6$ or $m = - 2$ .

CostGives the roots directly, no checking; needs a leading coefficient of 1 (or one that divides evenly).

The key step of the Vieta route is “eliminate the parameter, then make a product of two brackets equal to a constant”. That step is really adding a term, a factorization trick.

Stop 2 · Factor first if you can

The family

m x^{2} - (3 m + 2) x + 6 = 0

. Click through the cells one by one and watch the discriminant box: is it ever not a perfect square?

The discriminant of this family is $9 m^{2} - 12 m + 4 = (3 m - 2)^{2}$ : always a perfect square. Yet only four cells have a ✓.

Factor it and the reason is clear:

m x^{2} - (3 m + 2) x + 6 = (m x - 2) (x - 3),

so the roots are $3$ and $\frac{2}{m}$ . For $\frac{2}{m}$ to be an integer, $m$ must divide $2$ : $m = \pm 1, \pm 2$ . And the cell marked !: when $m = 0$ it is not a quadratic equation at all. If the problem says “quadratic equation”, $m = 0$ is excluded; if it only says “all roots of the equation are integers”, then $m = 0$ (root $3$ ) does count.

Stop 3 · Switch the main variable: solve for the parameter

If the parameter appears only to the first power, you can turn things round and treat it as the unknown.

Example: integers m for which x² − mx + 2m − 1 = 0 has integer roots

Step 1 / 5

Put the terms with m on one side
$m$ appears only to the first power. Rewrite the equation as “ $m$ times something equals something”.

$m (x - 2) = x^{2} - 1$
Solve for m and divide
Being an integer becomes divisibility
x = 3 or 5
x = 1 or −1

Stop 4 · Estimation: squeezed between two consecutive squares

Sometimes you have to prove that there are no integer roots. The trick is to squeeze $Δ$ between two consecutive perfect squares, with no other square in between.

The family

x^{2} + m x - (m + 2) = 0

. Read the small print under the discriminant: which two squares is it squeezed between?

Here $Δ = m^{2} + 4 m + 8 = (m + 2)^{2} + 4$ . When $m$ is a positive integer,

(m + 2)^{2} < Δ < (m + 3)^{2},

because $(m + 3)^{2} - (m + 2)^{2} = 2 m + 5 > 4$ . So $Δ$ is not a perfect square, and the equation has no integer roots (not even rational ones). The only ✓ in the whole row is at $m = - 2$ : there $Δ = 4$ and the equation is $x^{2} - 2 x = 0$ .

Practice

Exercise 1Basic

Both roots of $x^{2} - 7 x + c = 0$ are positive integers. What is the largest possible $c$ ?

Exercise 2Basic

Both roots of $x^{2} + m x + 6 = 0$ are integers. Give all possible values of $m$ .

Exercise 3Advanced

$m$ is an integer and both roots of $x^{2} - m x + m + 5 = 0$ are integers. Give all possible values of $m$ .

Exercise 4Advanced

$m$ is an integer, and both roots of the quadratic equation $m x^{2} - (m + 3) x + 3 = 0$ in $x$ are integers. Give all possible values of $m$ .

Exercise 5Challenge

$m$ is an integer and the equation $x^{2} - m x + 3 m - 2 = 0$ has integer roots. Give all possible values of $m$ .

In one sentence

The discriminant route: when does x² − mx + m + 2 = 0 have integer roots?

Stop 1 · Vieta's formulas: no discriminant needed

The discriminant route

The Vieta route

Stop 2 · Factor first if you can

Stop 3 · Switch the main variable: solve for the parameter

Example: integers m for which x² − mx + 2m − 1 = 0 has integer roots

Stop 4 · Estimation: squeezed between two consecutive squares

Practice

Mistake Museum · Exhibits from this stop

Thinking a perfect-square discriminant means integer roots

Not checking whether the equation is quadratic