Finding number of positive integral solutions of $x^4-y^4=3879108$ Find the number of positive integral solutions of $$x^4-y^4=3879108$$ $$3879108=36*277*389$$
I tried simplifying factors of $3879108$ to get terms in the form of $x^4-y^4$. However, I am unable to proceed. Do the (positive integral) solutions exist, and if yes, how to obtain them?
 A: A broad hint: first of all, it's clear that $x$ and $y$ must have the same parity (why?).  If $x$ and $y$ are both even, then $x^4$ and $y^4$ are both divisible by $16$, and so their difference must also be.  Suppose instead that $x$ and $y$ are both odd.  Now, $x^4 = \left(x^2\right)^2$, and likewise $y$; what do you know about squares of odd numbers?  (Alternately, as suggested in the comments: you can factor $x^4-y^4 = (x-y)(x+y)(x^2+y^2)$; now, look at the parity of the terms on the RHS.)
A: According to my understanding, this should be the answer.
$$LHS=x^4-y^4$$
For the value of this expression to be even $(3879108)$, both $x$ and $y$ should be either even or odd
$$LHS=(x+y)(x-y)(x^2+y^2)$$
Now, $(x+y)$, $(x-y)$, $(x^2+y^2)$ will all be even when $x$ and $y$ have the same parity. In other words, $$LHS=(2k)(2m)(2n)=8kmn$$ where k, m, n are some constants. Thus, $RHS$ must be divisible by $8$. However, $3879108$ is not divisible by $8$. Thus, $$x^4-y^4=3879108$$ has no positive integral solutions.
