How to prove that $50^{44}+30! \cdot44^{50}-24 $ is divisible by $31$?

Question

I am struggling with this example. This is a new topic for me. I would really appreciate some hints to understand it. I need part a) and b) to conclude the excercise.

I was thinking about Fermat, Euler or Wilson theorems but I am not sure if it is correct, of course I can use properties of congruences in number theory.

Part a): show that $50^{44}+30! \cdot 44^{50}-24 $ is divisible by $31$
Part b) : Conclude that $ (50^{44}+30!\cdot 44^{50})^{\varphi(31)} \equiv 1 \pmod{ 31}$

Answer 1

Here's a handful of hints:

$50^{44} = 2^{44}5^{88}$

$44^{50} = 2^{100}11^{50}$

$2^5\equiv 1 \pmod{31}$

$5^3 \equiv 1 \pmod{31}$

$11^2 \equiv -3 \pmod{31}$

Answer 2

\begin{align} 50^{44} + 30! \cdot 44^{50} &\equiv 50^{44} - 44^{50} \pmod{31}, \text{ By Wilson's Theorem } \\ &\equiv 2^{44} \cdot 5^{88} - 2^{100}\cdot 11^{50} \pmod{31} \\ &\equiv 2^{14}\cdot 5^{28} - 2^{10} \cdot 11^{20} \pmod{31}\text{, By Fermat's little Theorem } \\ &\equiv 2^{4} \cdot 5 - (-3)^{10} \pmod{31}, \\ &\equiv 2^4 \cdot 5 - 3^{10} \pmod{31} \\ &\equiv 2^4 \cdot 5 - 3^{3\times 3}\cdot 3 \pmod{31} \\ &\equiv 2^4 \cdot 5 - (-4)^3 \cdot 3 \pmod{31} \\ &\equiv 2^4 (5+12) \pmod{31} \\ &\equiv 2^4 (2^4+1) \pmod{31} \\ &\equiv2^8 + 2^4 \pmod{31} \\ &\equiv 2^3 + 2^4 \pmod{31} \\ &\equiv 24 \pmod{31} \end{align}

Answer 3

Notice we can use Fermat's little theorem to reduce the exponents using mod order reduction, based on these values: $\ \color{#0a0}{2^5\equiv 1},\,$ $\, \color{#c00}{5^3\equiv 1},\,$ $\, \color{#90f}{11^3}\equiv 11(-3) \equiv \color{#90f}{-2},\,$ yielding

$$\begin{align} \bmod 31\!:\quad\ & 50^{44}\!+30! \cdot 44^{50}\\[,2em] \equiv\ & 50^{14}\ \ \ \, -\,\ \ \ 44^{20}\qquad\ \ {\rm by}\ \ 30!\equiv -1\,\ {\rm by\ Wilson}\\[,2em] \equiv\ &\color{#0a0}{2^{14}} \cdot \color{#c00}{5^{28}}-\color{#0a0}{2^{40}} \cdot \color{#90f}{11^{20}}\\[.3em] \equiv\ &\ \ \ \ \ \frac{\color{#c00}{5}}{\color{#0a0}2}\,\ -\,\ \color{#0a0}1\cdot \color{#90f}{\frac{(-2)^{7}}{11} \ \ \ \text{by (*) below}}\\[.3em] \equiv\ &\ \ \ \ \frac{\color{}{36}}{2}\ \ \ \ \ +\ \ \ \ \ \frac{\color{#90f}{2^{2}}}{11}\quad\ \ {\rm by}\ \ \color{#0a0}{2^5\equiv 1}\\[.3em] \equiv\ &\ \ \ \ 18\ \ +\ \ \frac{\color{}{4+62}}{11}\\ \equiv\ &\ \ \ \ 24 \end{align}\!\!\!$$

$\color{#90f}{\text{(*)}}\,$ we used $\ \color{#90f}{11^3 = -2}\ \, \smash{\overset{\Large x^{7}\!}\Longrightarrow}\ 11^{21} \equiv (-2)^7\!\Rightarrow \color{#90f}{11^{20}\equiv (-2)^7/11},\ $ then in the next two lines we twiddle the fraction numerators by adding (or subtracting) small multiples of the modulus so to make the quotient exact.

Part b) is just FLT again.

Check here to see how modular fractions work.

Answer 4

Considering that

$\quad 2^5 \equiv 1 \pmod{31} \land 50 \equiv 19 \pmod{31} \land 44 \equiv 13 \pmod{31}$

we calculate (looking for 'presentation loop closure'),

$\quad 13^2 \equiv 2^1 \cdot 7^1 \pmod{31}$

$\quad 19^2 \equiv 2^2 \cdot 5^1 \pmod{31}$

$\quad 7^2 \equiv 2^1 \cdot 3^2 \pmod{31}$

$\quad 5^3 \equiv 1 \pmod{31}$

$\quad 3^5 \equiv 2^1 \cdot 13^1 \pmod{31}$

and write

\begin{align} 50^{44} + 30! \cdot 44^{50} &\equiv 19^{44} - 13^{50} \pmod{31}, \text{ By Wilson's Theorem } \\ &\equiv 2^{44} \cdot 5^{22} - 2^{25}\cdot 7^{25} \pmod{31} \\ &\equiv 2^{4} \cdot 5^{1} - 2^{12} \cdot 3^{24} \cdot 7 \pmod{31} \\ &\equiv 2^{4} \cdot 5^{1} - 2^{12} \cdot 2^{4} \cdot 13^{4} \cdot 3^4 \cdot 7 \pmod{31} \\ &\equiv 2^{4} \cdot 5^{1} - 2^{12} \cdot 2^{4} \cdot 2^2 \cdot 7^2 \cdot 3^4 \cdot 7 \pmod{31} \\ &\equiv 2^{4} \cdot 5^{1} - 2^{12} \cdot 2^{4} \cdot 2^2 \cdot 2 \cdot 3^2 \cdot 3^4 \cdot 7 \pmod{31} \\ &\equiv 2^{4} \cdot 5^{1} - 2^{4} \cdot 3^6 \cdot 7 \pmod{31} \\ &\equiv 2^{4} \cdot 5^{1} - 2^{4} \cdot 2^1 \cdot 13^1 \cdot 3 \cdot 7 \pmod{31} \\ &\equiv 2^{4} \cdot 5^{1} - 3 \cdot 7 \cdot 13 \pmod{31} \\ &\equiv 80 + 130 \pmod{31} \\ &\equiv 24 \pmod{31} \end{align}

Answer 5

This answer is Community wiki and takes a 'deep dive' into algorithmic solution design.

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The 'elemental' exponent reducing identity in ${\displaystyle (\mathbb {Z} /31\mathbb {Z} )^{\times}}$ is

$\quad \large 2^1\cdot3^1\cdot5^1 \equiv 2^0\cdot3^0\cdot5^0 \cdot (-1) \pmod{31}$

and this can be expanded into a 'closed exponent reducing presentation system',

$\quad \large 2^4 \equiv 3^1\cdot5^1\cdot (-1) \pmod{31}$
$\quad \large 3^3 \equiv 2^2 \cdot (-1) \pmod{31}$
$\quad \large 5^2 \equiv 2^1\cdot3^1\cdot (-1) \pmod{31}$

Part 1: Calculate $50^{44} \pmod{31}$

Since

$\quad 50 \equiv 19 \pmod{31}$

$\quad 19 \equiv 2^2 \cdot 3 \cdot (-1) \pmod{31}$

our 'not too smart' algorithm can be applied:

\begin{align} 50^{44} &\equiv 19^{44} \pmod{31} \\ &\equiv 2^{88} \cdot 3^{44} \cdot 5^0 \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{0} \cdot 3^{66} \cdot 5^{22} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{44} \cdot 3^{0} \cdot 5^{22} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{0} \cdot 3^{11} \cdot 5^{33} \cdot (-1)^{1} \pmod{31} \\ &\equiv 2^{0} \cdot 3^{11} \cdot 5^{33} \cdot (-1)^{1} \pmod{31} \\ &\equiv 2^{16} \cdot 3^{27} \cdot 5^{1} \cdot (-1)^{1} \pmod{31} \\ &\equiv 2^{34} \cdot 3^{0} \cdot 5^{1} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{2} \cdot 3^{8} \cdot 5^{9} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{6} \cdot 3^{12} \cdot 5^{1} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{14} \cdot 3^{0} \cdot 5^{1} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{2} \cdot 3^{3} \cdot 5^{4} \cdot (-1)^{1} \pmod{31} \\ &\equiv 2^{4} \cdot 3^{5} \cdot 5^{0} \cdot (-1)^{1} \pmod{31} \\ &\equiv 2^{6} \cdot 3^{2} \cdot 5^{0} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{2} \cdot 3^{3} \cdot 5^{1} \cdot (-1)^{1} \pmod{31} \\ &\equiv 2^{4} \cdot 3^{0} \cdot 5^{1} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{0} \cdot 3^{1} \cdot 5^{2} \cdot (-1)^{1} \pmod{31} \\ &\equiv 2^{1} \cdot 3^{2} \cdot 5^{0} \cdot (-1)^{0} \pmod{31} \\ &\equiv 18 \pmod{31} \end{align}

Part 2: Calculate $44^{50} \pmod{31}$

Since

$\quad 44 \equiv 13 \pmod{31}$

and

$\quad 13 \equiv 2 \cdot 3^2 \cdot (-1) \pmod{31}$

our 'not too smart' algorithm can be applied:

\begin{align} 44^{50} &\equiv 13^{50} \pmod{31} \\ &\equiv 2^{50} \cdot 3^{100} \cdot 5^0 \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{116} \cdot 3^{1} \cdot 5^{0} \cdot (-1)^{1} \pmod{31} \\ &\equiv 2^{0} \cdot 3^{30} \cdot 5^{29} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{20} \cdot 3^{0} \cdot 5^{29} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{34} \cdot 3^{14} \cdot 5^{1} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{2} \cdot 3^{22} \cdot 5^{9} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{16} \cdot 3^{1} \cdot 5^{9} \cdot (-1)^{1} \pmod{31} \\ &\equiv 2^{0} \cdot 3^{5} \cdot 5^{13} \cdot (-1)^{1} \pmod{31} \\ &\equiv 2^{6} \cdot 3^{11} \cdot 5^{1} \cdot (-1)^{1} \pmod{31} \\ &\equiv 2^{12} \cdot 3^{2} \cdot 5^{1} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{0} \cdot 3^{5} \cdot 5^{4} \cdot (-1)^{1} \pmod{31} \\ &\equiv 2^{2} \cdot 3^{2} \cdot 5^{4} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{4} \cdot 3^{4} \cdot 5^{0} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{0} \cdot 3^{5} \cdot 5^{1} \cdot (-1)^{1} \pmod{31} \\ &\equiv 2^{2} \cdot 3^{2} \cdot 5^{1} \cdot (-1)^{0} \pmod{31} \\ &\equiv 2^{1} \cdot 3^{1} \cdot 5^{0} \cdot (-1)^{1} \pmod{31} \\ &\equiv 25 \pmod{31} \end{align}