if $a_{1},a_{2},a_{3},a_{4}$ be rational numbers then find $a_{1}+a_{2}+a_{3}+a_{4}$

Question

Let $a_{1},a_{2},a_{3},a_{4}$ be rational numbers such that $$\{a_{i}a_{j}|1\le i<j\le 4\}=\{-24,-2,-\dfrac{3}{2},-\dfrac{1}{8},1,3\}$$ Find $a_{1}+a_{2}+a_{3}+a_{4}$

I was able to solve this problem with the following

$$(a_{1}a_{2})(a_{1}a_{3})\cdots (a_{3}a_{4})=(a_{1}a_{2}a_{3}a_{4})^3=(-24)(-2)(-\frac{3}{2})(-\frac{1}{8})(1)(3)=27$$ $$a_{1}a_{2}a_{3}a_{4}=3$$ Could you suggest a helpful idea with $a_{1}+a_{2}+a_{3}+a_{4}$

Answer 1

Building on the idea that $a_1 a_2 a_3 a_4 = 3$, assume WLOG that

$$\begin{align} a_1 a_2=1 \\ a_3 a_4=3 \end{align}$$

and further that

$$\begin{align} a_1=x,\quad a_2=\frac{1}{x},\quad |x|>1 \tag{1}\\ a_3=y,\quad a_4=\frac{3}{y},\quad |y|>\sqrt3 \tag{2} \end{align}$$

Exclude $|x|=1$ because this would mean $a_1 a_3 = a_2 a_3$ which is not allowed.

So then, the greatest magnitude amongst these possible pairwise products is clearly $xy$, and the smallest clearly $\frac{3}{xy}$, so:

$$\left.\begin{array}{c} xy=-24 \\ \frac{3}{xy}=-\frac{1}{8} \end{array}\right\} \implies \boxed{xy=-24} \tag{3}$$

Other possible pairwise products are $\dfrac{y}{x},\dfrac{3x}{y}$ and these are some permutation of $\{-2,-\frac{3}{2}\}$. So

$$\frac{3x/y}{y/x}=\frac{4}{3}\text{ or }\frac{3}{4} \implies 3\frac{x^2}{y^2}=\frac{4}{3}\text{ or }\frac{3}{4} \implies \frac{x}{y}=-\frac{2}{3}\text{ or }-\frac{1}{2} \tag{4}$$

Comparing (4) with (3):

$$y^2=\frac{xy}{x/y}=36\text{ or }48$$

Since $y$ is rational, $y=\pm6$.

$$y=-6,x=4 \implies a_1=4,a_2=\frac{1}{4},a_3=-6,a_4=-\frac{1}{2} \implies a_1+a_2+a_3+a_4=-\frac{9}{4}$$ or $$y=6,x=-4 \implies a_1=-4,a_2=-\frac{1}{4},a_3=6,a_4=\frac{1}{2} \implies a_1+a_2+a_3+a_4=\frac{9}{4}$$

So $$\boxed{a_1+a_2+a_3+a_4=\pm\frac{9}{4}}$$

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There may be a cleaner way, e.g. constructing a degree 4 polynomial with the required sum and product of roots. But it may be difficult to ensure that the pairwise products cover the entire set of six values and ensure that the roots are all rational.

Answer 2

You can now deduce, without loss of generality, $$\begin{align} a_1a_2&=-24&a_3a_4&=-\frac18\\ a_1a_3&=-2&a_2a_4&=-\frac32\\ \{a_1a_4,a_2a_3\}&=\left\{1,3\right\} \end{align}$$

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Suppose $a_2a_3=1$. Then the four equations above imply $$(a_1a_2)(a_1a_3)(a_2a_3)=-24(-2)(1)=48\implies a_1a_2a_3=\pm4\sqrt{3}$$ and this is incompatible with the numbers being rational.

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So $a_2a_3=3$. Then the four equations above imply $$(a_1a_2)(a_1a_3)(a_2a_3)=-24(-2)(3)=144\implies a_1a_2a_3=\pm12$$ and $a_4=\frac{a_1a_2a_3a_4}{a_1a_2a_3}=\pm\frac14$. This implies $$(a_1,a_2,a_3,a_4)=\left(\pm4,\mp6,\mp\frac12,\pm\frac14\right)$$ making $$a_1+a_2+a_3+a_4=\mp\frac94$$

Answer 3

Ignoring minus signs for the moment, we see that since all six products are distinct, one of them is $1$, and only one is less than $1$, the absolute values of the four numbers have the form $a$, $b$, $1/b$, and $c$ with

$$1/b\lt a\lt b\lt c$$

Since the smallest (absolute) product is $1/8$ and the largest is $24$, we conclude the four absolute values comprise the set

$$\left\{{1\over b},{b\over8},b,{24\over b}\right\}$$

with

$$\left\{{b^2\over8},{24\over b^2}\right\}=\left\{{3\over2},2\right\}$$

The only solution with $b$ rational is $b=4$, so the absolute values of the four numbers are $1/4$, $1/2$, $4$, and $6$. Finally, for two of the actual products to be postive and six negative, we must have two of the actual numbers for positive and two negative; since one of the positive products is $1$, the only possibilities are $(1/4,-1/2,4,-6)$ and $(-1/4,1/2,-4,6)$, with sums $-9/4$ and $9/4$ respectively.