Help with basic modular arithmetic question

Question

I asked this question moments ago but it was closed despite the link being non-applicable to my question. I need help solving the below basic modular problem,

Show that $39$ divides $17^{48}-5^{24}$

My course has NOT covered Fermat's Theorem. The only properties of modular arithmetic I know are additivity, multiplicativity, symmetry, and transitivity. My professor wants me to solve this question only using these basic properties, but I have honestly no idea how to do so. I don't know what pattern I'm supposed to recognize.

Answer 1

Well $\mod 3$ it's easy to see $17^{48} - 5^{24} \equiv (-1)^{48}-(-1)^{24} \equiv 0 \pmod 3$ so $3|17^{48}-5^{24}$.

And $\mod {13}$ we see $17^{48} - 5^{24} \equiv 4^{48} - 5^{24}\equiv 16^{24} - 5^{24} \equiv 3^{24}-5^{24}\pmod {13}$

Now... can't use Fermat (which makes short work) so we use sequential squaring.

$3^2\equiv 9$ and $3^3 \equiv 27 \equiv 1\pmod {13}$. And $5^2 \equiv 25 \equiv -1\pmod {13}$

So $3^{24} - 5^{24} \equiv 27^8 - 25^{12} \equiv 1^8 - (-1)^{12} \equiv 1 - 1 \equiv 0\pmod{13}$

So $13|17^{48} - 5^{24}$.

So $\operatorname{lcm}(3,13) = 39|17^{48} - 5^{24}$.

Answer 2

Those two exponents are very even (meaning have many factors of $2$), so let's apply difference of squares a few times, hoping that we can get a useful factor that is much smaller.

\begin{align*} 17^{48} - 5^{24} &= (17^{24} - 5^{12})(17^{24} + 5^{12}) \\ &= (17^{12} - 5^{6})(17^{12} + 5^{6})(17^{24} + 5^{12}) \\ &= (17^{6} - 5^{3})(17^{6} + 5^{3})(17^{12} + 5^{6})(17^{24} + 5^{12}) \end{align*}

$39 = 3 \cdot 13$. If $3$ divides any of these factors and $13$ divides any of these factors, $39$ divides the original quantity. If $3$ divides a quantity, that quantity is congruent to $0 \pmod{3}$. If $13$ divides a quantity, that quantity is congruent to $0 \pmod{13}$.

Let's compute $17^6 - 5^3 \pmod{3}$. \begin{align*} 17^6 &\cong (-1)^6 \pmod{3} \\ &\cong 1 \pmod{3} \text{ and } \\ 5^3 &\cong 1^3 \pmod{3} \\ &\cong 1 \pmod{3} \text{, so} \\ 17^6 - 5^3 &\cong 1 - 1 \pmod{3} \\ &\cong 0 \pmod{3} \text{.} \end{align*} Therefore, $3$ divides the first factor.

Now let's compute $17^6 - 5^3 \pmod{13}$. It is useful to note that $6 = 4+2 = 110_2$ in binary, so $17^6 = 17^{4+2} = 17^4 \cdot 17^2$ and we can get $17^4$ by squaring $17^2$. (For more on this, see exponentiation by squaring and its application to modular exponentiation.) \begin{align*} 17^2 &\cong (4)^2 \pmod{13} \\ &\cong 16 \pmod{13} \\ &\cong 3 \pmod{13} \text{ and } \\ 17^4 &\cong (17^2)^2 \pmod{13} \\ &\cong (3)^2 \pmod{13} \\ &\cong 9 \pmod{13} \text{ so } \\ 17^6 &\cong 17^4 \cdot 17^2 \pmod{13} \\ &\cong 9 \cdot 3 \pmod{13} \\ &\cong 27 \pmod{13} \\ &\cong 1 \pmod{13} \text{ and also } \\ 5^3 &\cong 5 \cdot 5 \cdot 5 \pmod{13} \\ &\cong 25 \cdot 5 \pmod{13} \\ &\cong (-1) \cdot 5 \pmod{13} \\ &\cong 8 \cdot 5 \pmod{13} \text{ so } \\ 17^6 - 5^3 &\cong 1 - 8 \pmod{13} \\ &\cong -7 \pmod{13} \\ &\cong 6 \pmod{13} \text{.} \end{align*}

So $13$ doesn't divide the first factor. How about the second factor? \begin{align*} 17^6 + 5^3 &\cong 1 + 8 \pmod{13} \\ &\cong 9 \pmod{13} \end{align*}

Nope. The third factor? \begin{align*} 17^{12} + 5^6 &\cong (17^6)^2 + (5^3)^2 \pmod{13} \\ &\cong 1^2 + 8^2 \pmod{13} \\ &\cong 65 \pmod{13} \\ &\cong 5 \cdot 13 \pmod{13} \\ &\cong 0 \pmod{13} \\ \end{align*}

So $13$ divides the third factor. Therefore, $39$ divides the given quantity.

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Here's another method that is more direct.

\begin{align*} 17^{48} - 5^{24} &\cong 17^{3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 } - 5^{3 \cdot 2 \cdot 2 \cdot 2} \pmod{39} \\ &\cong (17^3)^{2 \cdot 2 \cdot 2 \cdot 2 } - (5^3)^{2 \cdot 2 \cdot 2} \pmod{39} \\ &\cong (4913)^{2 \cdot 2 \cdot 2 \cdot 2 } - (125)^{2 \cdot 2 \cdot 2} \pmod{39} \\ &\cong (125\cdot 39 + 38)^{2 \cdot 2 \cdot 2 \cdot 2 } - (3 \cdot 39 + 8)^{2 \cdot 2 \cdot 2} \pmod{39} \\ &\cong (38)^{2 \cdot 2 \cdot 2 \cdot 2 } - (8)^{2 \cdot 2 \cdot 2} \pmod{39} & & \text{[alt.: note that $38 \cong -1 \pmod{39}$]} \\ &\cong (38^2)^{2 \cdot 2 \cdot 2 } - (8^2)^{2 \cdot 2} \pmod{39} \\ &\cong (37 \cdot 39 +1)^{2 \cdot 2 \cdot 2 } - (1 \cdot 39 + 25)^{2 \cdot 2} \pmod{39} \\ &\cong (1)^{2 \cdot 2 \cdot 2 } - (25)^{2 \cdot 2} \pmod{39} \\ &\cong 1 - (25)^{2 \cdot 2} \pmod{39} \\ &\cong 1 - (25^2)^{2} \pmod{39} \\ &\cong 1 - (16 \cdot 39 + 1)^{2} \pmod{39} &\cong 1 - (1)^{2} \pmod{39} \\ &\cong 0 \pmod{39} \text{,} \end{align*} so $39$ divides the gven quantity.

Answer 3

I'll provide two proofs. The second one is probaly the one you're looking for

$17^{48}-5^{24}=(17^{16})^3-(5^{4})^6(mod 39)$

By Euler Totient Theorem, you have thats equal to

$1^3-1^6=1-1=0(mod 39)$, Thus, it is divisible by 39

Second proof:

$17^{48}=(17^2)^{24}=(-1)^{24}=1 (mod39) $

$5^{24}=(5^3)^8=8^8=(8^4)^2=1^2=1 (mod 39)$

Together, you get $1-1=0 (mod 39)$. Thus, it is proven