1965 AHSME Problems/Problem 40
Problem
Let be the number of integer values of such that is the square of an integer. Then is:
Solution
First, we wish to factor into a more manageable form. From the beginning of , we notice , which gives us the idea to use .
This gives us This is not useful, but it gives us a place to start from.
We can then try . This is much more useful, as it moves all non-linear terms inside of a squared expression.
We can then say , where is the square of an integer mentioned on the problem. Right from here, we can set , which cancels out the , giving . This gives us one solution, , .
We can then rearrange the expression, giving us . Factoring using difference of squares, we obtain
We can then state that when is greater than and less than , will be greater than . This is obtained by setting and then solving the inequality. We can then conclude that .
Next, we claim that or when . We can prove this by first noting that since and are integers, is an integer. Next, we shall assume that . Solving this and plugging back into the original equation, we obtain . Solving we obtain , which is a contraction to . Therefore, and or .
Finally, we can go back to the equation We note that since is larger than , in order for there to be solutions, must be in the range . However, this contradicts what was proven earlier, so when and or , there are no solutions for .
Now, all that remains to be checked are values of between and . Using brute force and checking each value individually, we can assert that there are no such solutions for , leaving us with only solution, . Therefore, the answer is
Solution by treetor10145
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
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