1967 AHSME Problems/Problem 31
Problem
Let , where , , are consecutive integers and . Then is:
Solution
Let . Then , which simplifies to .
From the options, we want to test if is always a perfect square. Because the polynomial expression for is quartic, if it is a perfect square, it would be the square of a quadratic expression. Thus, could be written in the form for some .
Setting , we can compare coefficients. From the coefficient, we get . Note that if works, so does , so we can arbitrarily pick .
We now have . Setting the cubic terms equal gives , or . This leaves . We can quickly inspect the constant term to determine that . We reject , since the quadratic and linear terms won't match up, which leaves as the only possibility - and, in fact, it works.
Thus, is always the square of an integer - namely . This in turn means that is always rational, which leaves choices as the only possible correct answers.
The question now is whether , or , is odd, even, or could be both. We have two cases for :
If , then . This means .
If , then , and .
Either way, is an odd integer, and the answer is
See also
1967 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 30 |
Followed by Problem 32 | |
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