Let $x=\frac{a+kb}{2}$ ofcurse $x$ is natural then this can be written as $x^2-kxb+b^2+1$ Wlog $max (x,b)=x$ then: If $x=b$ then we esily have $k=3$ and $x=b=1$ Now let $(x_0,b_0)$ be the smaller sollution then there is a second sollution $(x_1,b_0)$ with: $x_1+x_0=kb_0$ and $x_1x_0=b^2+1$ from this two we get $x_1$ natural. Also sinse $(x_0,b_0)$ is the smaller sollution then $x_1>=x_0$ but we have: $b_0^2+1=x_1*x_0\geqslant x_0^2\geqslant (b_0+1)^2>b_0^2+1$ contradiction

just look to case $a$ divisible by 2 and not ,and use if $a^2+b^2$ divisible by $p=4k+3$ where $p$ is prime $\rightarrow$ $a$ divisble by p and $b$

P2nisic wrote: Let $x=\frac{a+kb}{2}$ ofcurse $x$ is natural then this can be written as $x^2-kxb+b^2+1$ Wlog $max (x,b)=x$ then: If $x=b$ then we esily have $k=3$ and $x=b=1$ Now let $(x_0,b_0)$ be the smaller sollution then there is a second sollution $(x_1,b_0)$ with: $x_1+x_0=kb_0$ and $x_1x_0=b^2+1$ from this two we get $x_1$ natural. Also sinse $(x_0,b_0)$ is the smaller sollution then $x_1>=x_0$ but we have: $b_0^2+1=x_1*x_0\geqslant x_0^2\geqslant (b_0+1)^2>b_0^2+1$ contradiction Doesn't seem to quite work as written but does lead to the famous Vieta jumping finish once the nice sub is made.