51st Ukrainian National Mathematical Olympiad, 4th Round

We will prove this by contradiction. Denote the sum $x^c+y^d$ by $s$. Then, obviously, $x^c=-y^d(\mathrm{mod}s)$ and also if $(x^a+y^b):s$, then $x^a=-y^b(\mathrm{mod}s)$. This implies that $x^{ad}=(-1{)}^d{y}^{bd}(\mathrm{mod}s)$ and $x^{bc}=(-1{)}^b{y}^{bd}(\mathrm{mod}s)$, hence $(-1{)}^d{x}^{ad}=y^{bd}=(-1{)}^b{x}^{bc}(\mathrm{mod}s)\Rightarrow x^{ad}=(-1{)}^{b-d}{x}^{bc}(\mathrm{mod}s)$. It is clear that $x$ and $s$ are coprime, so we can divide the last congruence by $x$ raised to the power $\min \{ad,bc\}$, and obtain $x^{\max \{ad,bc\}}=(-1{)}^{b-d}(\mathrm{mod}s)$. Similarly, we get $y^{\max \{ad,bc\}}=(-1{)}^{a-c}(\mathrm{mod}s)$. Therefore, $y^{\max \{ad-bc\}}-x^{\max \{ad-bc\}}:s$ or $y^{\max \{ad-bc\}}+x^{\max \{ad-bc\}}:s$. By the statement of the problem, we have $0<|ad-bc|<\min \{c,d\}$. But then $$|y^{\max \{ad-bc\}}-x^{\max \{ad-bc\}}|<y^{\max \{ad-bc\}}+x^{\max \{ad-bc\}}<y^d+x^c=s.$$ So, the expression $|y^{\max \{ad-bc\}}\pm x^{\max \{ad-bc\}}|$ can be divisible by $s$ only if it is equal to zero. But this contradicts the fact that $x$ and $y$ are coprime, and we are done.