1961 AHSME Problems/Problem 33
Problem
The number of solutions of , in which and are integers, is:
Solution
Let and . Substituting these values results in Factor the difference of squares to get If , then , so can not be negative. If , then . Since is always positive, the result would be way less than , so can not be negative. Thus, and have to be nonnegative, so and are integers. Thus, From the first case, and . Since does not have an integral solution, the first case does not work. From the second case, and . Thus, and . Thus, there is only one solution, which is answer choice .
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 32 |
Followed by Problem 34 | |
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