equation $x^2+ax+6a = 0$ has integer roots, Then integer values of $a$ is

Question

If the equation $x^2+ax+6a = 0$ has integer roots, Then integer values of $a$ is

$\bf{My\; Try}::$ Let $\bf{\alpha,\beta}$ be two roots of given equation $x^2+ax+6a = 0$

So $\bf{\alpha+\beta = -a}$ and $\bf{\alpha \cdot \beta = 6a}$ and $\bf{\alpha,\beta \in \mathbb{Z}}$

So $\bf{\alpha \cdot \beta =-6\alpha -6\beta \Rightarrow 6\alpha+\alpha \cdot \beta +6\beta +36 = 36}$

So $\bf{(\alpha+6)\cdot (\beta+6) = 36 = 6\times 6 = 9\times 4 = 18\times 2 = 36\times 1}$ and many more

But I did not understand how can i calculate for all ordered pairs.

Is there is any other method to solve it

Help Required

Thanks

@juantheron: There are $18$ integer solutions. Hint, write the expression for the solutions in terms of $a$ and then plot them. Hopefully, that will provide you a hint. — Amzoti, Dec 26 '13 at 18:24
A differetn way to arrive at $(\alpha+6)(\beta+6)=36$ is by observing that $ax$ cancels against $6a$ if $x=-6$ — Hagen von Eitzen, Dec 26 '13 at 18:34
@Amzoti I fond only $8$ solutions (see my answer below). Where did you get your $18$ from? — Hagen von Eitzen, Dec 26 '13 at 18:36
@Amzoti $3$ and $8$ won't work. Any positive solution to $a$ has to be at least $24$. — Josephine Moeller, Dec 26 '13 at 18:49

Hagen von Eitzen · Answer 1 · 2013-12-26T18:31:53.300

You've done nothing wrong so far. And there's not "and many more", for we may assume wlog. that $\alpha\ge \beta$ and then your enumeration is complete at least for the cases with $\alpha+6,\beta+6\ge 0$. The same factorings with negative numbers are of course possible as well. All in all you get that $$\begin{align}a=-\alpha-\beta&\in\{0+0,-3+2,-12+4,-30+5, 15+10, 24+8, 42+7\}\\&=\{0,-1,-8,-25,25,32,49\}\end{align}$$ and in fact all these are correct solutions.

EDIT: Oops, your "and many more" indeed was justified - slightly - as you had left out the factorization $12\times 3$. This adds $-6+3=-3$ and $18+9=27$ to the list of solutions.

You've missed out $a=24$. When $a=24$, we have
$$x^2+24x+144 \equiv (x+12)(x+12)$$ — Fly by Night, Dec 27 '13 at 17:52

score 2 · Answer 2 · answered Jun 11 '17 at 04:26

What you have done so far is absolutely correct. However, for the sake of completeness I will repeat the beginning of your solution. Using Vieta's equations, we know that $$\alpha+\beta=-a$$ and $$\alpha\beta=6a$$

By substituting the first equation into the second, we get: $$ \alpha\beta=-6\alpha-6\beta$$ Which simplifies to:

$$(\alpha+6)(\beta+6)=36$$ Since we are only interested in $a$ a solution $(x,y)$ would be the same as a solution $(y,x)$, hence, we assume without loss of generality that $\alpha \leq \beta$. $$ $$ Hence, we know that $$(\alpha+6,\beta+6)=(1,36),(2,18),(3,12),(4,9),(6,6),(-36,-1),(-18,-2),(-12,-3),(-9,-4),(-6,-6)$$

Which give us our values for $\alpha$ and $\beta$: $$(\alpha,\beta)=(-5,30),(-4,12),(-3,6),(-2,3),(0,0),(-42,-7),(-24,-8),(-18,-9),(-15,-10),(-12,-12)$$

Therefore, the possible values for $a$ are: $$a=-25, -8, -3, -1, 0, 49, 32, 27, 25, 24$$

However, I do not think that this was your problem in solving the question. As indicated in your question, we were unsure how to calculate all ordered pairs. The key to solving problem was realizing that there were only a small number of ways to express 36 as the product of 2 integers. In fact, there were only 10 ways to do this. This is true for most small integers. When in doubt as to the feasibility of testing all cases, it can be useful to calculate the number of factors that the integer has. This can be done by breaking the number down into its prime factorization. For example, $36=2^2 \cdot 3^2$, to find the number of positive factors, we simply find the number of ways to separate $2^2 \cdot 3^2$ into 2 groups. There are 3 possibilities for 2, we can either have $2^0$ or $2^1$ or $2^2$ in the first group, each of which will produce unique factors, similarly, there are 3 possibilities for 3. Hence, the number of positive factors of $36$ is $3 \cdot 3=9$ As such, the total number of factors that $36$ has is $18$ (including negative factors). The number of ways to express 36 as the product of $2$ integers would, under normal circumstances, be half this, as each factor should have a unique other factor with which it can multiply to form $36$, (e.g. $1 \cdot 36=36$), however, 36 is a perfect square and has to repeated factors($6\cdot 6$ and $-6 \cdot -6$). Hence, the number of ways 2 express $36$ as the product of $2$ integers is $\frac{9\cdot 2 -2}{2} +2=10$, which is quite feasible to test.

score 2 · Answer 3 · answered Dec 26 '13 at 18:21

2

The two roots are $x_{1,2}=\dfrac{-a\pm\sqrt{a^2-24a}}{2}\in\mathbb{Z}\iff\sqrt{a^2-24a}\in\mathbb{N}\iff a(a-24)=n^2$. Also, the numerator has to be even. Obviously, $a=0$ and $a=25$ are two such solutions.

answered Dec 26 '13 at 18:21

Lucian

48,334
2
83
154

score 2 · Answer 4 · answered Dec 26 '13 at 18:48

2

Hint $\ $ Scaling by $-4$ and completing the square yields

$$ 144 = A^2 - X^2 = (A-X)(A+X)\ \ {\rm for}\ \ A = 12-a,\ \ X = 2x+a$$

so it reduces to the finite problem of testing various factorizations of $144$.

answered Dec 26 '13 at 18:48

Bill Dubuque

272,048

This solution admits several fractional solutions that don't really need to be checked. Setting $A = 6 - a/2$ and $X = x + a/2$ and factoring $36$ instead of $144$ eliminates a lot of checking. – Josephine Moeller Dec 26 '13 at 19:11
@John Not if one notices that the factors must have the same parity. The point is that this is a special case of a general way of solving bivariate diophantine equations of this form, see here. – Bill Dubuque Dec 26 '13 at 19:23

equation $x^2+ax+6a = 0$ has integer roots, Then integer values of $a$ is

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