First $a>b$ now note that we can bound $p$ by taking minimal values of $a,b$ (here we take $(a,b)=(3,2)$ because $(a,b)=(2,1)$ doesnt yold sols) thus we have $3^7-2^7=2059>2013$ thus $p=2,3,5$. But note that $3^p=2013+2^p$ also doesnt yeld solutions becuase $p=5$ its just too small and $p=7$ its too big then the next bound i take $(a,b)=(5,4)$ becuase $(a,b)=(4,3),(4,2),(4,1)$ give that $p=5$ its just too small and $p=7$ way bigger. By this bound we have that $5^5-4^5=2101>2013$ thus $p=2,3$.(Thus works becuase small $a,b$ gives bigger $p$ and i just did a massive case worm that i'm too lazy to write lol). Now assume that $p=3$, then if there exists any sol with $3 \mid a-b$ then by LTE $$1=v_3(2013)=v_3(a^3-b^3)=v_3(a-b)+1 \ge 2 \; \text{contradiction!!}$$Meaning that $3$ will never divide $a-b$ meaning that $3$ cant divie both $a,b$ thus by FLT: $$0 \equiv 2013=a^3-b^3 \equiv a-b \not\equiv 0 \pmod 3 \; \text{contradiction!!}$$Thus $p=2$ and then the equations becomes $(a-b)(a+b)=2013$ and note that $2013=3 \cdot 11 \cdot 61$ thus we have this cases: $a-b=3$ and $a+b=671$ thus $(a,b)=(337,334)$ $a-b=11$ and $a+b=183$ thus $(a,b)=(97,86)$ $a-b=33$ and $a+b=61$ thus $(a,b)=(47,14)$ Thus all the solutions are: $(337,334,2), (97,86,2), (47,14,2)$ And we are done Man this problem sucks, its such a waste of time doing a lot of casework for bounding $p$ :c. I'm really tired and i'm doing this at mid-night so good night...