Square-Free Numbers: Is $k^k + K - 1$ Always Square-Free?

Is the number $k^k + k - 1$ always square-free? This is a fascinating question in elementary number theory that has intrigued mathematicians and number enthusiasts alike. A square-free number, as you guys probably know, is an integer that is not divisible by any perfect square other than 1. For example, 10 is square-free because its prime factorization is 2 × 5, and neither 2 nor 5 appear more than once. On the other hand, 12 is not square-free because it's divisible by 4 (which is 2 squared). So, let's get into the thick of it and explore this question in detail.

Initial Observations and Testing

Okay, so the initial question is: For what values of k is $k^k + k - 1$ a square-free number? Our user has already done some groundwork by testing values up to $k = 17$ and hasn't found any counterexamples. This is a great start! It gives us some empirical evidence to suggest that perhaps $k^k + k - 1$ is indeed square-free for all positive integers k. But, as we all know, a few examples don't make a proof. We need to dig deeper.

When tackling a problem like this, it's always good to start with small values and see if any patterns emerge. Let's test a few more values of k and observe the results:

• For $k = 1$, $1^1 + 1 - 1 = 1$, which is square-free.
• For $k = 2$, $2^2 + 2 - 1 = 5$, which is square-free.
• For $k = 3$, $3^3 + 3 - 1 = 29$, which is square-free.
• For $k = 4$, $4^4 + 4 - 1 = 259$, which is square-free.
• For $k = 5$, $5^5 + 5 - 1 = 3129$, which is square-free.

So far, so good. But, we need a more systematic approach. Just checking numbers one by one isn't going to cut it. We need to think about what it means for a number to be square-free and how we can prove that a number of the form $k^k + k - 1$ satisfies this condition.

Diving into Number Theory

To really get a handle on this problem, we need to bring in some tools from number theory. Remember, a number is square-free if it is not divisible by any perfect square (other than 1). This means that if $k^k + k - 1$ is not square-free, then there exists some prime number p such that $p^2$ divides $k^k + k - 1$. This is the key idea we need to explore.

Let's assume, for the sake of contradiction, that there is a prime $p$ such that $p^2$ divides $k^k + k - 1$. This can be written mathematically as:

$k^k + k - 1 ≡ 0 ext{ (mod } p^2)$

This congruence relation is crucial because it tells us that $k^k + k - 1$ leaves a remainder of 0 when divided by $p^2$. Our goal now is to try and manipulate this congruence to see if we can derive any contradictions or useful information about k and p.

One approach might be to consider the expression modulo p as well. If $p^2$ divides $k^k + k - 1$, then p must also divide $k^k + k - 1$. So we also have:

$k^k + k - 1 ≡ 0 ext{ (mod } p)$

This gives us two congruences to work with, which is often a good strategy in number theory problems. Now, the challenge is to find a way to link these congruences and see if we can find any restrictions on p and k. This is where things can get tricky, and we might need to explore different techniques and theorems from number theory.

Exploring Congruences and Modular Arithmetic

Alright, let's delve a bit deeper into modular arithmetic and see if we can extract some more information from our congruences. We have two key congruences to work with:

1. $k^k + k - 1 ≡ 0 ext{ (mod } p^2)$
2. $k^k + k - 1 ≡ 0 ext{ (mod } p)$

The second congruence, $k^k + k - 1 ≡ 0 ext{ (mod } p)$, tells us that $k^k + k - 1$ is divisible by p. This is a good starting point. We can rewrite this congruence as:

$k^k ≡ 1 - k ext{ (mod } p)$

This form might be helpful because it isolates $k^k$ on one side. Now, let's think about what we can do with this. One common technique in number theory is to raise both sides of a congruence to some power. However, we need to be careful about what power we choose. We want to choose a power that might simplify the expression or reveal some hidden structure.

Another avenue to explore is to consider the order of k modulo p. The order of an integer k modulo p (denoted as ord_p(k)) is the smallest positive integer n such that $k^n ≡ 1 ext{ (mod } p)$. The order can be a powerful tool because it tells us about the powers of k modulo p. If we can find the order of k modulo p, it might help us simplify the congruence $k^k ≡ 1 - k ext{ (mod } p)$.

Lifting the Exponent Lemma (LTE) – A Potential Tool

Given that we are dealing with powers and congruences, a potentially useful tool here might be the Lifting The Exponent Lemma (LTE). LTE is a powerful result that helps us determine the highest power of a prime p that divides $x^n - y^n$ under certain conditions. While our expression is of the form $k^k + k - 1$, we might be able to massage it into a form where LTE can be applied.

LTE generally comes in handy when we have expressions like $x^n - y^n$ and we want to find the largest power of a prime p that divides it. The lemma has different forms depending on whether p is odd or even, and whether p divides x - y. The most common form is for odd primes p:

If p is an odd prime, and x and y are integers such that $p | (x - y)$ but p does not divide x and y, then

$v_p(x^n - y^n) = v_p(x - y) + v_p(n)$

where $v_p(n)$ is the exponent of the highest power of p that divides n. To consider the applicability of LTE, we need to relate our expression $k^k + k - 1$ to something of the form $x^n - y^n$. This might involve some algebraic manipulation or clever substitutions.

Considering Specific Cases and Potential Contradictions

Let's step back and think about what we're trying to achieve. We're assuming that $k^k + k - 1$ is not square-free, meaning there exists a prime p such that $p^2$ divides $k^k + k - 1$. Our goal is to either find such a p and k, or to show that no such p and k can exist, leading to a contradiction.

One way to approach this is to consider specific cases. For example, what if p is a small prime like 2 or 3? Can we find any values of k such that $k^k + k - 1$ is divisible by 4 or 9? If we can rule out small primes, it might give us a better handle on the general case.

Another approach is to try and find some bounds on p. If we can show that p must be greater than some value, it might simplify the problem. For example, if we could show that p must be greater than k, that would be a significant step forward.

To make progress, let's consider the case where $p = 2$. If $p^2 = 4$ divides $k^k + k - 1$, then $k^k + k - 1 ≡ 0 ext{ (mod } 4)$. Let's examine the possible values of k modulo 4:

• If $k ≡ 0 ext{ (mod } 4)$, then $k^k + k - 1 ≡ 0^0 + 0 - 1 ≡ -1 ≡ 3 ext{ (mod } 4)$
• If $k ≡ 1 ext{ (mod } 4)$, then $k^k + k - 1 ≡ 1^1 + 1 - 1 ≡ 1 ext{ (mod } 4)$
• If $k ≡ 2 ext{ (mod } 4)$, then $k^k + k - 1 ≡ 2^2 + 2 - 1 ≡ 4 + 2 - 1 ≡ 5 ≡ 1 ext{ (mod } 4)$
• If $k ≡ 3 ext{ (mod } 4)$, then $k^k + k - 1 ≡ 3^3 + 3 - 1 ≡ 27 + 3 - 1 ≡ 29 ≡ 1 ext{ (mod } 4)$

In none of these cases is $k^k + k - 1$ congruent to 0 modulo 4. Therefore, 4 cannot divide $k^k + k - 1$ for any k. This is a good start! We've ruled out p = 2.

The Road Ahead: More Exploration and Potential Proof Techniques

So, we've made some progress. We've tested some small values of k, ruled out p = 2, and explored some potential tools like LTE and modular arithmetic. But we're not there yet. This problem seems to require a more intricate approach. Here's a recap of what we've done and some possible directions for future exploration:

• Recap:
  • We're trying to determine if $k^k + k - 1$ is always square-free.
  • We've tested values up to k = 17 and found no counterexamples.
  • We've established the congruences $k^k + k - 1 ≡ 0 ext{ (mod } p^2)$ and $k^k + k - 1 ≡ 0 ext{ (mod } p)$.
  • We've ruled out the case p = 2.
• Next Steps:
  • Explore other small primes: We should investigate the cases p = 3, 5, 7, and so on, to see if we can find any patterns or contradictions.
  • Refine bounds on p: Can we establish any lower or upper bounds on p in terms of k? This could help narrow down the possibilities.
  • Apply LTE more strategically: We need to find a way to express $k^k + k - 1$ (or a related expression) in a form suitable for LTE. This might involve some clever algebraic manipulation.
  • Consider the order of k modulo p: The order of k modulo p could provide valuable information about the powers of k and their relationships modulo p.
  • Look for contradictions: The core strategy is to assume that $k^k + k - 1$ is not square-free and try to derive a contradiction. This might involve showing that certain conditions on p and k cannot be simultaneously satisfied.

This problem is a classic example of how number theory can present deceptively simple questions that require deep and creative thinking to solve. It's likely that a complete solution will involve a combination of the techniques we've discussed, along with perhaps some new insights and approaches. Keep experimenting, keep exploring, and most importantly, keep having fun with the math!

Conclusion

So, is $k^k + k - 1$ always square-free? We don't have a definitive answer yet, but we've explored the problem in depth, discussed various approaches, and even ruled out one case (p = 2). The journey through this problem highlights the beauty and challenge of number theory. It's a field where seemingly simple questions can lead to complex and fascinating investigations. Whether the answer is ultimately yes or no, the process of trying to solve this problem will undoubtedly teach us a lot about the intricacies of numbers and their properties. Keep up the great work, guys, and who knows, maybe one of you will be the one to crack this problem wide open!