The Sum of Two Squares and the Intractable Case of 7^k-1

Is there a positive integer k such that 7^k-1 is the sum of two squares? Let’s explore this intriguing question through the lens of number theory.

Introduction to Quadratic Residues

Before diving into the specific problem, it's important to understand the concept of quadratic residues. A quadratic residue modulo n is any number that is congruent to a square modulo n. In simpler terms, a number is a quadratic residue if it can be expressed as x^2 modulo n. For example, the quadratic residues modulo 8 are 0, 1, and 4, as these are the only residues that can be obtained by squaring integers modulo 8.

Understanding quadratic residues is crucial because they help us determine whether a number can be represented as the sum of two squares. According to a well-known result in number theory, a number can be written as the sum of two squares if and only if in its prime factorization, every prime of the form 4k 3 appears an even number of times.

The Intractable Case: 7^k-1

Let's now consider the specific case of 7^k-1. Our goal is to prove that for any positive integer k, 7^k-1 cannot be the sum of two squares.

Case 1: k is Odd

Suppose k is odd. Then, we have the following congruence:

7^k ≡ (-1)^k ≡ -1 (mod 8)

Since k is odd, -1 ≡ 7 (mod 8), so:

7^k-1 ≡ 6 (mod 8)

However, no number congruent to 6 modulo 8 can be expressed as the sum of two squares, as the possible quadratic residues modulo 8 are 0, 1, and 4. No two of these numbers add up to 6. Hence, if k is odd, 7^k-1 cannot be the sum of two squares.

Case 2: k is Even

Now, let's consider the case where k is even. We can write k as 2a, where a is a positive integer. Therefore:

7^k-1 7^(2a) - 1 (7^a-1)(7^a 1)

Assume, for the sake of contradiction, that 7^k-1 is the sum of two squares. This means that 7^a-1 must not be the sum of two squares (as k is the smallest integer for which 7^k-1 is a sum of two squares). By the aforementioned result in number theory, 7^a-1 must have a prime factor p, and p must be of the form 4k 3 and appear an odd number of times.

However, the difference between 7^a 1 and 7^a-1 is 2. This means that if 7^a-1 is divisible by a prime p of the form 4k 3, then 7^a 1 must also be divisible by the same prime p (since 2 is not divisible by any such prime). This is a contradiction because no prime of the form 4k 3 can divide both 7^a-1 and 7^a 1, as their difference is 2.

Thus, 7^a-1 and 7^a 1 cannot both be divisible by a prime of the form 4k 3, and therefore, their product (7^k-1) cannot be a sum of two squares. This is a contradiction to our assumption, proving that no positive integer k exists such that 7^k-1 is the sum of two squares.

What's the Issue?

The issue arises from the fact that while 7^k-1 is always divisible by 6 (which includes the prime factor 3), the additional prime factors (like 3 and 19 in the case of 7^k-1 for certain values of k) play a crucial role. For instance:

7^1-1 is divisible by 2 (2^1) and 3 (3^1), and thus, fails to be a sum of two squares. 7^3-1 is divisible by 3^2 (9) and 19 (19^1), but 9 is not a problem, whereas the 19 prevents it from being a sum of two squares. The key takeaway is that for higher values of k, the prime factorization of 7^k-1 introduces primes of the form 4k 3 in a way that ensures the formation is not a sum of two squares.

This observation highlights the complexity of the problem and the subtleties involved in prime factorization and congruences.

Conclusion

Through a combination of modular arithmetic and prime factorization, we have shown that there is no positive integer k such that 7^k-1 is the sum of two squares. This problem serves as a fascinating illustrative case in number theory, showcasing the intricate interplay between primes, congruences, and quadratic residues.