Difference between revisions of "1991 AIME Problems/Problem 8"
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<cmath>\begin{eqnarray*}sr + 6s + 6r &=& 0\\ | <cmath>\begin{eqnarray*}sr + 6s + 6r &=& 0\\ | ||
sr + 6s + 6r + 36 &=& 36\\ | sr + 6s + 6r + 36 &=& 36\\ | ||
− | (s + 6)(r + 6) &=& 36</cmath> | + | (s + 6)(r + 6) &=& 36 |
+ | \end{eqnarray*} | ||
+ | </cmath> | ||
[[Without loss of generality]] let <math>r \le s</math>. | [[Without loss of generality]] let <math>r \le s</math>. |
Latest revision as of 18:40, 10 March 2015
Problem
For how many real numbers does the quadratic equation have only integer roots for ?
Solution
Solution 1
By Vieta's formulas, where are the roots of the quadratic, and since are integers, must be an integer. Applying the quadratic formula,
Since is an integer, we need to be an integer (let this be ): . Completing the square, we get
Which implies that is a perfect square also (let this be ). Then
The pairs of factors of are ; since is the average of each respective pair and is also an integer, the pairs that work must have the same parity. Thus we get pairs (counting positive and negative) of factors that work, and substituting them backwards show that they all work.
Solution 2
Let . Vieta's yields .
Without loss of generality let .
The possible values of are: .
See also
1991 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.