This question was uploaded on 26/01/22 on social media accounts.
Solution:
Given equation:
\[\color{blue} {a^2+2ab-3b^2-41=0}\]
\[\Rightarrow \color{blue} {a^2\color{red}{+a^2-a^2}+2ab-3b^2-41=0}\]
\[\Rightarrow \color{blue} {2a^2-(a^2-2ab+b^2)-2b^2-41=0}\]
\[\Rightarrow \color{blue} {2(a^2-b^2)-(a-b)^2-41=0}\]
\[\Rightarrow \color{blue} {2(a+b)(a-b)-(a-b)^2=41}\]
\[\Rightarrow \color{blue} {(a-b)(a+3b)=1\times41}\]
Given that a and b belong to natural numbers. So, both must be positive integers.
Clearly, ( a - b ) < ( a + 3b )
\[\color{red} {a-b=1}\Rightarrow\color{red} {a=1+b}\]
Now put this value in other equation:
\[\color{red} {a+3b=41}\Rightarrow\color{red} {1+4b=41}\]
\[\Rightarrow\color{red} {b=10}\]
\[\Rightarrow\color{red} {a=11}\]
\[\color{fuchsia} {a^2+b^2=11^2+10^2=221}\]
