I have to summon Fermat again

First, as $7$ is prime, by Fermat's Little Theorem $n^{7} \equiv n \pmod{7} \Longrightarrow n^{7} - n \equiv 0 \pmod{7}$ . So $n^{7} - n$ is divisible by $7$ for all positive integers $n$ .

Next, note that $n^{7} - n = n(n^{6} - 1) = n(n^{3} - 1)(n^{3} + 1) = n(n - 1)(n^{2} + n + 1)(n + 1)(n^{2} - n + 1)$ .

Now since the factors $n - 1, n$ and $n + 1$ are three consecutive integers we know that precisely one of them is divisible by $3$ , and also that at least one of them is divisible by $2$ , and so $n^{7} - n$ will be divisible by $3 \times 2 = 6$ .

Thus as $n^{7} - n$ is divisible by both $7$ and $6$ it will be divisible by $7 \times 6 = 42$ for all positive integers $n$ . But is this the maximum possible value $M$ ?

Now $2^{7} - 2 = 126 = 42 \times 3 = 14 \times 9$ , so we need to check out if $M$ could equal $126$ . But

$3^{7} - 3 = 3 \times 3^{6} - 3 = 3 \times 9^{3} - 3 \equiv 0 - 3 \pmod{9} \equiv 6 \pmod 9$ ,

i.e., we have at least one case where $n^{7} - n$ is not divisible by $9$ , and hence also not divisible by $126$ . Thus we can conclude that $M = \boxed{42}$ .

I'm attempting to show the following more general result: For $N>1$ , the largest $M$ such that $n^N-n$ is divisible by $M$ for all positive $n$ is $M=\prod_{(p-1)|(N-1)}p$ , where $p$ is a prime. In the case of $N=7$ we find $M=2\times3\times7=\boxed{42}$ .

For a prime $p$ , we have $n^N\equiv n \pmod{p}$ iff $n^{N-1}\equiv 1 \pmod{p}$ for all $n$ that are not divisible by $p$ . Fermat tells us that this is the case iff $p-1|N-1$ . Note that $M$ is squarefree since $p^N-p$ fails to be divisible by $p^2$ .

Note that 5 does not necessarily divide $n^7-n$ since $2^7-2 = 126$ is not divisible by $5$ .

By Fermat's Little Theorem , $n^7\equiv n\bmod 7$ and thus $n^7 - n$ is divisible by $7$ . Now, factor: $n^7-n = (n-1)n(n+1)(n^2 - n + 1)(n^2 + n + 1)$ Since $n-1$ , $n$ , and $n+1$ are three consecutive integers, at least one must be even and one must be divisible by $3$ . Thus both $2$ and $3$ divide $n^7-n$ as well. Now, let $p>7$ be prime and let $n = 2$ $\begin{aligned} n^7-n &= (n-1)n(n+1)(n^2 - n + 1)(n^2 + n + 1) \\ &= 1\times 2\times 3\times 3\times 7 \\ &\not\equiv 0 \mod p &&\color{#3D99F6} \text{Since }2,3,7<p \end{aligned}$ Thus $n^7-n$ is not divisible by $p$ for any prime greater than $7$ .

At this point, we know that $M=2^a3^b7^c$ . We now show that $a=b=c=1$ . We do this by showing that, with the proper choice of $n$ , $n^7-n$ is not divisible by $p^2$ for prime $p$ :

Let $n = p$ . Then $\begin{aligned} n^7-n &= (n-1)n(n+1)(n^2 - n + 1)(n^2 + n + 1) \\ &= (p-1) p (p+1) (-p+1) (p+1) \\ &\not\equiv 0\mod p^2 && \color{#3D99F6}\text{Since none of }p-1, p+1, -p+1 \text{ are divisible by }p\text{, }p\text{ appears only once as a factor} \end{aligned}$

Therefore, $M=2\times 3\times 7 = \boxed{42}$ .

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As an alternative to the last step, one could also use $n=42$ . Then $n^7-n = 230,539,333,206$ which is not divisible by $4$ , $9$ or $49$ .

Let f(n) = n^7 - n.

We start with n=2.

What ist the max. factor of f(2)? It is 7.

gcd( f(2), f(3), f(4), f(5), f(6) ) = 42 =M

$2^7-2=126$ , so M must be a divisor of $126=2×3^2×7$ .

Also $3^7-3=2184$ , so M must also be a divisor of $2184=2^3×3×7×13$ .

Combined we now know that M must be a divisor of $2×3×7=42$ , we only need to prove that 42 works indeed. Using modular arithmetic, we see that for any p: $0^7-0≡0$ , $1^7-1≡0$ and $(-1)^7-(-1)≡0$ (mod p). And Fermats little theorem ensures that $n^7-n=0$ (mod 7). The facts $n^7-n≡0$ (mod 2), $n^7-n≡0$ (mod 3), and $n^7-n≡0$ (mod 7) combine into $n^7-n≡0$ (mod 42).

Without invoking Fermat's Little Theorem: A simple experiment with a spreadsheet showed that for n=2 then $n^{7}-n$ = 126 which is 3 x 42 whereas if n=3 then $n^{7}-n$ = 2184 which is 52 x 42. Given that 3 and 52 are mutually prime the number M we are seeking cannot be greater than 42. So, is it 42?

Factorising as much as I can gives $n^{7}-n = n(n^{6}-1) = n(n^{3}-1)(n^{3}+1) = n(n-1)(n^{2}+n+1)(n^{3}+1)$

It is clear that either n or n-1 is divisible by 2.

Working mod 3, then either n $\equiv$ 0 or n $\equiv$ 1 (in which case n-1 $\equiv$ 0) or n $\equiv$ 2 (in which case $n^{3}+1 \equiv 8+1 \equiv 2+1 \equiv$ 0)

Working mod 7 needs a little more effort:

$n \equiv\ 0$ implies $n \equiv\ 0$

$n \equiv\ 1$ implies $n-1 \equiv\ 0$

$n \equiv\ 2$ implies $n^{2}+n+1 \equiv 4+2+1 \equiv\ 0$

$n \equiv\ 3$ implies $n^{3}+1 \equiv 27+1 \equiv 6+1 \equiv\ 0$

$n \equiv\ 4$ implies $n^{2}+n+1 \equiv 16+4+1 \equiv 2+4+1 \equiv\ 0$

$n \equiv\ 5$ implies $n^{3}+1 \equiv 125+1 \equiv 6+1 \equiv\ 0$

$n \equiv\ 6$ implies $n^{3}+1 \equiv 216+1 \equiv 6+1 \equiv\ 0$

So M is a multiple of 2, 3 and 7 and not greater than 42. Hence $M = \boxed{42}$

Not sure if this has already been proposed. Here's an alternate solution, albeit tedious but more general, using binomials. Though not as elegant as Fermat's Little Theorem approach, it is a neat little property of binomials.

Any polynomial function $f(n)$ of degree $d$ over any field that contains $\mathbb{Q}$ can be expressed uniquely as the following linear combination, $f(n) = \sum_{k=0}^d F_k {n\choose k}$ where $F_k=\sum_{t=0}^k(-1)^{k-t}f(t){k\choose t}$ is the k-th difference of the sequence $\{f(i)\}_{i=0,2,\ldots,k}$ .

Solution to problem

Let $f(n) = n^7 - n$ and so $d=7$ , we can express it as $f(n) = 126{n\choose 2} + 1806{n\choose 3} + 8400{n\choose 4} + 16800{n\choose 5} + 15120{n\choose 6} + 5040{n\choose 7}.$ The GCD of the coefficients in the equation is $\boxed{42}$ .

By Fermat's Little Theorem, we have $n^p=n \pmod p$ for all $n$ and primes $p$ , in particular for $p=2, 3, 5, 7$ .

Thus, 7 divides $n^7-n$ .

Also 2 divides it, because $n^7 = n \cdot n^2\cdot n^2\cdot n^2=n\cdot n\cdot n\cdot n = \ldots = n \pmod 2$ .

Similarly, 3 divides it, since $n^7=n\cdot n^3\cdot n^3=n\cdot n\cdot n=n \pmod 3$ .

For $p=5$ , this doesn't work, and also not for primes $p\gt5$ .

Hence, the largest $\fbox{\$M=2\cdot 3\cdot 7=42\$}$ .

I tried every even number and then divide the one I got for the first one which 126 then when I tried 6 the answer wasn’t divisible by 6 so I tried 42 bcuz it is a factor of 126 and it actually works

42 is just always the answer

Just take the the case of 2 and 3 and find the common factors which are in both and multiply them

(1) n^7-n=(n^3-n)(n^4+n^2+1), so it divides by 3.

(2) By Fermat's theorem, it divides by 7.

(3) It is even, it divides by 2.

Therefore, it divides by 2x3x7,

Answer=42

Since (M) is the largest value that evenly divides $n^7-n$ then $n^7-n , (n+1)^7-(n+1) , \cdots (n+k)^7 -(n+k)$ must share same gcd. The $\text{g.c.d} \,(n^7-n, (n+1)^7-(n+1) , (n+2)^7-(n+2) ,\cdots (n+k)^7-(n+k) )= 2\times 3\times 7 = 42$ And hence the answer.