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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.