1985 AHSME Problems/Problem 21
Problem
How many integers satisfy the equation
Solution
We recall that for real numbers and , there are exactly ways in which we can have , namely ; and ; or and is an even integer.
The first case therefore gives
Similarly, the second case gives , i.e. , and this indeed gives , so is a further valid solution.
Lastly, for the third case, we have but would give , which is odd, whereas gives , which is even. Therefore, this case gives only one further solution, namely .
Accordingly, the possible values of , , , or , yielding a total of solutions.
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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