First we'll factor 1998: $ 1998 = 2\cdot3^3\cdot37$ Let $ g = (x,y) = 2^a3^b37^c, x' = \frac {x}{g}, y' = \frac {y}{g}$ We have: $ {x'}^7 + {y'}^7 = 2^{z - 7a}3^{3z - 7b}37^{z - 7c}$ Lemma: If $ p^k||x^q + y^q, (p,q - 1), (x,y) = (p - 1,q) = (p,q) = 1$, $ p,q$ are primes: we must have $ p^k||x + y$. Proof: If k=0, this is trivial, because $ x + y|x^q + y^q$ when $ q \neq 2$, and when $ q = 2$: $ x^2 + y^2 \equiv x + y \equiv 1 \pmod 2$ If k>0, we have: $ ( - xy^{ - 1})^q \equiv 1 \pmod p, ( - xy^{ - 1})^{p - 1} \equiv 1 \pmod p$ $ \implies ( - xy^{ - 1})^{(p - 1,q)} \equiv - xy^{ - 1} \equiv 1 \pmod p$, or: $ x + y \equiv 0 \pmod p$. It's well-known that $ (x,y,p) = 1, p^{{k_1} > 0}||x + y, p \nmid n \implies p^{k_1}||x^n + y^n$ (follows directly from the identity $ (p^k\cdot t + y)^q - y^q \equiv tqy^{q - 1} p^k \pmod {p^{k + 1}}$, and can be verified separately for $ p = 2$) The lemma follows. In our problem, $ (x,y) = (x',y') = 1, q = 7, p = 2,3,37$. The identites $ (p - 1,q) = (p,q) = 1$ can be verified. So we have: $ 2^{z - 7a}3^{3z - 7b}37^{z - 7c} | x' + y'$. We conclude that $ |x'^7 + y'^7| \le |x' + y'|$. $ x',y' > 0 \implies x' = y' = 1$ which yields $ z - 7a = 1,3z = 7b,z = 7c$, which is a contradiction because it implies that $ z \equiv 0 (7)$ and also $ z \equiv 1 (7)$. So there are no solutions in positive integers. EDIT: generalization - The only possible solutions in integers to $ x^n + y^n = m^z$, where n,m are given positive integers such that $ (m\varphi(m),n) = 1$ are when $ x =y$ or $ xy=0$.