Proving Multiples in Exponents

$17^{7} - 17 = 17(17^{6} - 1) = 17((17^{2})^{3} - 1) = 17(17^{2} - 1)(17^{4} + 17^{2} + 1)$ ,

where the identity $a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})$ was used with $a = 17^{2}$ and $b = 1$ .

Now $17^{2} - 1 = (17 - 1)(17 + 1) = 16 \times 18 = 8 \times 36$ . Thus the answer is $\boxed{\text{Yes}}$ .

$\begin{aligned} 17^7 - 17 & \equiv 17(18-1)^6 - 17 \equiv 17(-1)^6-17 \equiv 17-17 \equiv 0 \text{ (mod 36)} \end{aligned}$ . Yes , $17^7-17$ is a multiple of 36.

Let $\lambda(\cdot)$ denote the Carmichael's lambda function , then $\lambda(36) = \lambda(2^2\cdot3^2) = \text{lcm}(\lambda(2^2), \lambda(3^2)) = \text{lcm}( \phi(2^2), \phi(3^2)) = 6 ,$

where $\text{lcm}$ denotes the lowest common multiple function, and $\phi(\cdot)$ denotes the Euler's totient function , with $\phi(2^2) = 2^2 \left(1 - \dfrac12\right) = 2$ and $\phi(3^2) = 3^2 \left( 1 - \dfrac13\right) = 6$ .

Knowing that $a^{\lambda(n)} \equiv 1 \pmod n$ for coprime positive integers $a,n$ , then

$\begin{aligned} 17^6 &\equiv &1 \pmod {36} \\ 17^7 &\equiv &17 \pmod {36} \\ \end{aligned}$

Hence, $17^{7} - 17$ is divisible by 36.

Relevant wiki: Chinese Remainder Theorem

By the Chinese Remainder Theorem, if $17^7 - 17$ is congruent to 0 modulo 9 and 4, then it is congruent to 0 modulo 36 as well, implying that $17^7 - 17$ is a multiple of 36. Since

$17^7 - 17 \equiv (-1)^7 - (-1) \equiv 0 \! \! \! \! \pmod{9}$

and

$17^7 - 17 \equiv 1^7 - 1 \equiv 0 \! \! \! \! \pmod{4},$

we conclude that $17^7 - 17 \equiv 0 \pmod{36}$ , and $\boxed{\text{Yes}}$ , it is a multiple of 36.

$17^0 = 1 \mod 36$
$17^1 = 17 \mod 36$
$17^2 = 1 \mod 36$

Because numbers raised to higher powers always follow cyclical patterns mod N, we know that the next terms of 17^n will be 17,1,17,1,17. Therefore, $17^7 = 17 \mod 36$ .

$17- 17 = 0$

First mod 4:

$17^7-17 \equiv 1^7 - 1 = 0 \pmod{4}$ .

Then mod 9:

$17^7-17 \equiv (-1)^7+1 = 0 \pmod{9}$ .

Thus 4 and 9 both divide $17^7-17$ , hence (since they are relatively prime) 36 = 4*9 also divides it.

Correct answer is: YES.

$17^{7} -17=17(17^{6}-1)=17(17^{3}+1)(17^{3}-1)$

Since $17^{3}$ is odd, both $17^{3}-1$ and $17^{3}+1$ are even.

Now:

$17^{3}+1=(18-1)^{3}+1$

All the terms on the cubic expansion are multiples of 18 except for the last one $(-1)^{3}=-1$ which cancels with the +1 (in fact:

$(18-1)^{3}=18^{3}-3.18^{2}+3.18-1$ )

So as one of the terms is multiple of 18 and the other is even, the product is multiple of 36.

Bonus

For two consecutive even numbers one is multiple of 4, so the result must bem multiple of 72

Bonus 2

$(18-1)^{3}+1=18^{3}-3.18^{2}+3^{18}-1+1=18(18^{2}-3.18+3)$

So this term must be a multiple of $18 \times 3$ , and the result mus have another factor of 3, making the number multiple of $72 \times 3=216$

$n=17^7-17=17(17^6-1)=17(17^3-1)(17^3+1)$ . Both $17^3-1$ and $17^3+1$ are even; hence, $n$ is divisible by $4$ . Since $17$ is not divisible by $3$ , exactly one of these factors is divisible by $3$ . But that is not enough.

Consider $17^3\bmod 9=(-1)^3\bmod 9=-1$ . Hence, $17^3+1$ is divisible by $9$ . Therefore, $n$ is divisible by $4\times 9=36$ .

Since 36 = 4 \cdot 9, if the expression is divisible by both 4 and 9, we are done. First we take mod 4:

17^7 - 7 \equiv (1)^7 - 1 \equiv 0 \pmod{4}

Next, mod 9:

17^7 - 7 \equiv (-1)^7 - (-1) \equiv 0 \pmod{9}

Thus the answer is YES .

Evaluate modulo 4: $17^7 - 17 \equiv 1^7 - 1 \equiv 1 - 1 \equiv 0 \ \ \text{mod 4}.$ Evaluate modulo 9: $17^7 - 17 \equiv (-1)^7 - (-1) \equiv (-1) - (-1) \equiv 0 \ \ \text{mod 9}.$ It follows that $17^7 - 17$ is a multiple both of 4 and 9, and therefore of their least common multiple, 36.

$\left . \begin{aligned} 17 \equiv 1 \pmod {4} \implies\ 17^6 \equiv 1 \pmod {4} \implies 4 \mid (17^6 - 1) \\ 17 \equiv -1 \pmod {9} \implies\ 17^6 \equiv 1 \pmod {9} \implies 9 \mid (17^6 - 1) \end {aligned} \right \} \implies 36 \mid (17^6 - 1) \implies 36 \mid (17^7 - 17)$

$\begin{aligned} { 17 }^{ 7 }-17 & = 17\left( { 17 }^{ 6 }-1 \right) \\ \quad & = 17\left( { 17 }^{ 3 }-1 \right) \left( { 17 }^{ 3 }+1 \right) \\ \quad & = 17\left( 17-1 \right) \left( { 17 }^{ 2 }+17+1 \right) \left( 17+1 \right) \left( { 17 }^{ 2 }-17+1 \right) \\ \quad & = 17\left( 16 \right) \left( 18 \right) \left( { 17 }^{ 4 }+{ 17 }^{ 2 }+1 \right) \\ { 17 }^{ 7 }-17 & = \left[ 17\left( 8 \right) \left( { 17 }^{ 4 }+{ 17 }^{ 2 }+1 \right) \right] \left( 36 \right) \end{aligned}$

By testmanship, yes is the likely answer as it would make no sense to ask if the answer was no.

Without using algebra higher other than “the difference of 2 squares or cubes”, in general the statement holds true for X^Y - X when X is odd and Y is odd. (Apologies for using ^ for "raising to the power"; I haven't figured out formatting yet).

e.g. For X is odd and Y = 3, Substitute N=X+1, and the expression must be divisible by 2N.

Rearranging and taking the difference of squares:

(N-1)^3 - (N-1) = (N-1) [(N-1)^2 - 1] = (N-1) x [(N-2) x N]

Since N is even (and X is odd) then (N-2) is also even (and so divisible by 2), so the product [(N-2) x N]) is divisible by 2N.
QED.

For Y = 5, we get

(N-1)^5 - (N-1) = N-1 x [(N-1)^4 x [(N-1^2 -1], where [(N-1)^2 - 1)], is the only relevant factor, the same as that above.

Any odd value of Y can be factored this way, where only the factor [(N-2)xN] is relevant:

(N-1)^Y - (N-1) = (N-1) x [N-1]^(Y-1) x [(N-2) x N)]

However, whenever Y is even, e.g. 4, we have to take the difference of 2 cubes

(N-1)^4 - (N-1) = (N-1) x [(N-1)^3 - 1] = (N-1) x [(N-1)^2 - (N-1) + 1)] =

(N-1) x [(N-1)^2 - N)] which is not In general divisible by 2N (Seems obvious but I realize I have not provided a formal proof here)