Suppose that $k_1k_2=\alpha$ and that $k_1+k_2=\gamma$. Are $k_1$ and $k_2$ unique?

Question

Let $k_1, k_2, \alpha, \gamma \in \mathbb{Z}$.

Suppose that $k_1k_2=\alpha$ and that $k_1+k_2=\gamma$.

Are $k_1$ and $k_2$ unique? That is, can one find another pair of integers whose sum is also $\alpha$ and whose product is also $\gamma$?

If so, how can one prove this?

Answer 1

Suppose you have

$$k_1k_2=\alpha \tag{1}\label{eq1A}$$

$$k_1+k_2=\gamma \tag{2}\label{eq2A}$$

WLOG, assume $k_1 \le k_2$. Assume there's a $k_3$ and $k_4$ where $k_3 \le k_4$, $k_3 \neq k_1$ such that

$$k_3+k_4=\gamma \tag{3}\label{eq3A}$$

There's some $d \neq 0$ where

$$k_3 = k_1 + d \tag{4}\label{eq4A}$$

Then you get

$$\begin{equation}\begin{aligned} k_4 & = \gamma - k_3 \\ k_4 & = k_1 + k_2 - (k_1 + d) \\ k_4 & = k_2 - d \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

You also have

$$k_3k_4 = (k_1 + d)(k_2 - d) = k_1k_2 - d^2 = \alpha - d^2 \tag{6}\label{eq6A}$$

This shows that, apart from order, the values of $k_1$ and $k_2$ are unique.

Answer 2

Yes they are unique up to a swapping operation.

By means of the identity

$$ k_1 k_2 = [ (k_1+k_2)/2]^2-[ (k_1-k_2)/2]^2 $$

which is the same as

$$\alpha = \gamma^2/4-[ (k_1-k_2)/2]^2 $$

we see that $(k_1,k_2)$ can be interchanged for same $\alpha,\gamma.$

If the sign of $\alpha $ is changed, then either one of $k_1$ and $k_2$ can undergo a sign change.

For example for product $+25$ if $(k_1,k_2)=(1,25)$ or $ (25,1)$ the sum is unchanged at $26$ by associative law.

Similarly for product $-25$ if $(k_1,k_2)=(-1,25)$ or $ (-25,1)$ the sum is unchanged at $-12.$

It can be also shown geometrically in Mohr's Circle of stress/strain.