# Difference between revisions of "1971 Canadian MO Problems/Problem 4"

## Problem

Determine all real numbers $a$ such that the two polynomials $x^2+ax+1$ and $x^2+x+a$ have at least one root in common.

## Solutions

### Solution 1

Let this root be $r$. Then we have $\begin{matrix} r^2 + ar + 1 &=& r^2 + r + a\\ ar + 1 &=& r + a\\ (a-1)r &=& (a-1)\end{matrix}$

Now, if $a = 1$, then we're done, since this satisfies the problem's conditions. If $a \neq 1$, then we can divide both sides by $(a - 1)$ to obtain $r = 1$. Substituting this value into the first polynomial gives $\begin{matrix} 1 + a + 1 &=& 0\\ a &=& -2 \end{matrix}$

It is easy to see that this value works for the second polynomial as well.

Therefore the only possible values of $a$ are $1$ and $-2$. Q.E.D.

### Solution 2

Let $x^2+ax+1 = (x-s)(x-t)$ and $x^2+x+a = (x-s)(x-t)$ where $s$ is the common root. From Vieta's Formulas, we have: $-(s+t) = a, -(s+u) = 1, st = 1,$ and $su = 1$. We see that $s,t,u \neq 0$. Dividing $su$ by $st$, we have: $$\frac{su}{st} = \frac{a}{1} \Rightarrow u = at$$ Also, we have: $$a+t = -s = 1+u \Rightarrow a+t = 1+u$$ Substituting $u = at$ into the above, we have: $\begin{matrix} a+t &=& 1+ at\\ at - a - t +1 &=& 0\\ (a-1)(t-1) &=& 0 \end{matrix}$

Thus either $a = 1$ or $t = 1$. We check to see that $a = 1$ is indeed a possible value to satisfy the requirements. If $t = 1$, then from $st = 1$, we have $s = t = 1$, and from $-(s+t) = a$, we have $a = -(1+1) = -2$, which also satisfies the requirements.

Thus, the only possible a values are: $a = 1, -2$.

## See Also

 1971 Canadian MO (Problems) Preceded byProblem 3 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • Followed byProblem 5
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