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

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In summary: Expert SummarizerIn summary, the maximum value of $a^2+b^2$ is 1982, which can be achieved by setting $a=1981$ and $b=1$. This is found by solving the equation $(a^2-ab-b^2)^2=1$ and considering the possible divisors of $b^2+1$.
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Determine the maximum value of $a^2+b^2$, where $a$ and $b$ are integers in the range $1,\,2,\,\cdots,\,1981$ satisfying $(a^2-ab-b^2)^2=1$.
 
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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.
 

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

1. What is the significance of maximizing $a^2+b^2$ for $(a^2-ab-b^2)^2=1$ in the context of "1981 Ints"?

The problem of maximizing $a^2+b^2$ for $(a^2-ab-b^2)^2=1$ is a well-known problem in mathematics, often referred to as the "1981 Ints" problem. It was first proposed in the International Mathematical Olympiad in 1981 and has since been a popular problem in various mathematical competitions. The solution to this problem involves finding the maximum value of $a^2+b^2$ for the given constraint, which has important implications in the field of number theory and algebraic geometry.

2. What is the approach to solving the "1981 Ints" problem?

The key to solving the "1981 Ints" problem is to use the concept of Vieta's formulas, which relate the coefficients of a polynomial to its roots. By applying Vieta's formulas to the given equation $(a^2-ab-b^2)^2=1$, we can express $a^2+b^2$ in terms of the roots of the polynomial. This allows us to find the maximum value of $a^2+b^2$ by finding the roots that satisfy the given constraint.

3. How does the solution to the "1981 Ints" problem relate to Pythagorean triples?

The solution to the "1981 Ints" problem involves finding integer solutions to the equation $(a^2-ab-b^2)^2=1$, which is similar to finding Pythagorean triples, where $a^2+b^2=c^2$. In fact, the solutions to the "1981 Ints" problem can be used to generate Pythagorean triples, making it a useful tool in number theory.

4. Are there any real-world applications of the "1981 Ints" problem?

While the "1981 Ints" problem may seem like a purely mathematical problem, it has real-world applications in fields such as cryptography and coding theory. The solutions to this problem can be used to generate secure codes and ciphers, making it an important problem in the field of data security.

5. Is there a generalization of the "1981 Ints" problem?

Yes, the "1981 Ints" problem can be generalized to finding the maximum value of $a^2+b^2$ for any polynomial of the form $(a^2-ab-b^2)^n=1$, where $n$ is a positive integer. This generalization has its own set of solutions and implications in mathematics, and is a popular problem in mathematical competitions as well.

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