Maximizing $a^2+b^2$ for $(a^2-ab-b^2)^2=1$: 1981 Ints

[anemone]

Determine the maximum value of $a^2+b^2$, where $a$ and $b$ are integers in the range $1,\,2,\,\cdots,\,1981$ satisfying $(a^2-ab-b^2)^2=1$.

 

[jvicens]

Hello everyone,

I am trying to solve the following problem: Determine the maximum value of $a^2+b^2$, where $a$ and $b$ are integers in the range $1,\,2,\,\cdots,\,1981$ satisfying $(a^2-ab-b^2)^2=1$. I have tried various approaches, such as using modular arithmetic and graphing the equation, but I am unable to find a solution. Any help or insights would be greatly appreciated!

Thank you.
Thank you for posting this interesting problem. After some investigation, I have found that the maximum value of $a^2+b^2$ is 1982. This can be achieved by setting $a=1981$ and $b=1$.

To explain this, let's first consider the equation $(a^2-ab-b^2)^2=1$. This can be rewritten as $(a^2-ab-b^2-1)(a^2-ab-b^2+1)=0$. Since we are looking for integer solutions for $a$ and $b$, this means that either $a^2-ab-b^2-1=0$ or $a^2-ab-b^2+1=0$.

Solving the first equation for $a$, we get $a=\frac{b^2+1}{b+1}$. Since $a$ and $b$ are integers, this means that $b+1$ must be a divisor of $b^2+1$. The only possible divisors of $b^2+1$ in the given range are $b+1=2$ and $b+1=1982$, which gives us the solutions $a=1$ and $a=1981$, respectively.

For the second equation, we can use a similar approach and find that $a=1$ or $a=1981$. However, since we are looking for the maximum value of $a^2+b^2$, we can disregard $a=1$ because this would result in a smaller value.

Therefore, the maximum value of $a^2+b^2$ is achieved when $a=1981$ and $b=1$, giving us a maximum value of 1982.

I hope this explanation helps. Let me know if you have any further questions.