First of all, must be distinct primes, if not, let's say , then a contradiction. Hence WLOG let .

Now just consider the last two conditions, since , the two conditions are equivalent to , note that if we treat as a constant, then is definitely boundable. In fact . Since the second term is less than 0, hence or .

Hence our equation is now . (tip: since we are dealing with primes, a natural instinct is to look at parity to see if could be used, in this case however it didn't spark anything). Now take a look at our first condition: , we take mod of our equation: , this means , solving this using factoring gives

the reason that I wrote the tip is because I actually considered it and ended up incorrectly deduce that , which I noticed while writing this proof. When I was working on this problem, I found that before the bound on , knowing this immediately gives you

As you can see with this example, although the conditions given in the problem are symmetric, you sometimes have to "ruin" it by taking it apart in order to solve it.