Solve in Z :
Let :
Solve this fourth degree polynomial (I can't do this yet so just assume you solve this for )
And you find that , , , .
If is negative, then , which is not possible, so if , it is not a solution in . The last solution is positive, but when squared does not give a relative number, so it is not a solution in .
Substitute back the only potential solution into ( ) to find out that the only solution in is :
Conclusion : admits one solution in : .
Another approach :
Note that the only perfect squares under are , , and . Denote any of these squares. If , , then is a solution for .
If , then we get a perfect square, thus the only solution in for your equation, is .
This is far simpler than my other fully algebraïc method, but it is conceptually harder to grasp, so it needs to be written a little better. Any tips ?