Does it help? $ y | (x^{2} + 1)$ $ k \cdot y =x^{2} + 1$ with $ k$ positive integer $ k \cdot y -1=x^{2}$ $ jky -j=jx^{2}$ $ x^{2} | (y^{3} + 1)$ $ j x^2 =y^{3} + 1$ with $ j$ positive integer $ jky -j=y^{3} + 1$ $ jky -(j+1)=y^3$ $ j+1=y(y^2-jk)$

Interesting! Since $ k=\frac{x^2+1}{y}$ and $ j=\frac{y^3+1}{x^2}$, we thus deduce $ kj =\frac{(x^2+1)(y^3+1)}{yx^2} =(1+\frac{1}{x^2})(y^2+\frac{1}{y}) > y^2$ Thus, $ y^2 - jk < 0$ From this $ j<-1$ which is impossible.

mszew wrote: $ j + 1 = y(y^2 - jk)$ should be $ j + 1 = y(jk-y^2)$

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