# Diego Marques

## A Challenge For You!!!!

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.

### Problem of the month (May 2015)

Find all solutions of the Diophantine equation

$$3^x-5^y=2.$$

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 $$y=2$$.$$\square$$

### Problem of the month (July 2015)

Find all positive integers $$n$$ such that

$$\alpha^n-n^2\alpha$$
is integer. Here $$\alpha$$ denotes the golden number: $$(1+\sqrt{5})/2$$.

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

$$\alpha^n-y^t\alpha$$
which is related to the equation $$F_n=y^t$$ and the only solutions are $$F_1=1^t$$, $$F_6=2^3$$ and $$F_{12}=12^2$$.