As a problem solver, I love to propose interesting problems to see new and ingenuous ideas. Since you clicked here I will propose you one problem. It may be easy or not, but I must say that none deep tools are necessary.
Find all solutions of the Diophantine equation
Solution: First, we look at the equation \(\pmod 3\), then we have \(5^y\equiv -2\equiv 1\pmod 3\) which yields \(y\) even and so \(y\geq 2\). Then \(x\geq 3\). By looking \(\pmod 9\) we have \(5^y\equiv -2\equiv 7\pmod 9\). However, \(5^y\equiv 5,7,8,4,2,1\pmod 9\) and then \(y\equiv 2\pmod 6\). Now, looking \(\pmod 7\), we obtain \(3^x\equiv 6\pmod 7\). However, \(3^x\equiv 3,2,6,4,5,1\pmod 7\) and then \(x\equiv 3\pmod 6\). By using that \(3^6\equiv 5^6\equiv 7\pmod{19}\), we have \(3^x\equiv 18,12,8\pmod{19}\) while \(5^y+2\equiv 6,11,8\pmod{19}\). Thus \(x\equiv 3\pmod{18}\) and \(y\equiv 2\pmod{18}\). By supposing \(x\geq 4\), we have \(5^y\equiv -2\equiv 79\pmod{81}\) leading to \(y\equiv 20\pmod{54}\). Since \(5^{54}\equiv 1\pmod{109}\), one has \(3^x\equiv 5^{20}+2\equiv 37\pmod{109}\), but on the other hand for \(x\equiv 3\pmod{18}\), it holds that \(3^x\equiv 16,66,27\pmod{109}\) yielding a contradiction. Thus the only solution happens with \(x=3\) and then \(y=2\).\(\square\)
Find all positive integers \(n\) such that
Solution: It suffices to use the well-known identity \(\alpha^n=\alpha F_n+F_{n-1}\). Then we want that \(\alpha (F_n-n^2)+F_{n-1}\) be an integer. Since \(\alpha\) is irrational, then \(F_n=n^2\). Now, we use that \(F_n\geq \alpha^{n-2}\) to obtain the that \(n\leq 12\). By testing the values we obtain that \(n=1\) and \(12\) are the only solutions. We point out that by using a theorem of Bugeaud, Mignotte and Siksek, one can think about the general case
