If x⁴-y⁴=15, then find the value of x⁴+y⁴, where x and y are natural numbers. · ABC26GN1228

$x^{4}-y^{4}=15 \Rightarrow\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right)=15=1 \times 15$ or $3 \times 5$ $\Rightarrow\left(x^{2}-y^{2}\right)=3$ and $\left(x^{2}+y^{2}\right)=5$ [ $\because x$ and $y$ are natural numbers] $\Rightarrow 2 x^{2}=8 \Rightarrow x^{2}=4 \Rightarrow x=2$. [Adding above two equations] Putting $x=2$ in $x^{2}+y^{2}=5$, we get: $y=1$. $\therefore \quad\left(x^{4}+y^{4}\right)=\left(2^{4}+1^{4}\right)=(16+1)=17$.