Prove that $x^{13}-y^{14}-x+y$ is divisble by $157$.

Question

Natural numbers $x,y$ are such that $x^{12}-y^{13}$ is divisible by $157$. Prove that $x^{13}-y^{14}-x+y$ is also divisble by $157$.

Off the cuff I tried to multiply first expression by some quantities to get to the $x^{13}-y^{14}-x+y$, but that would demand from $x^{12}-y^{13}$ to contain $\pm 1$ to obtain terms $x$ or $y$.

The key fact to me seems that $157$ is prime and can be expressed as $12 \times 13 +1$.

I checked by hand, in search of patterns, few cases in more general statement $n^2+n+1|x^n-y^{n+1} \rightarrow n^2+n+1|x^{n+1}-y^{n+2}-x+y$ and for $n=1,2$ it seems true.

Answer 1

It’s true if $x$ or $y$ is 0 mod 157, so we can assume both $x$ and $y$ are invertible mod 157. By the key fact you identified, along with the fundamental theorem of finite abelian groups, the group $k^*$ of units in the field $k=\mathbf{Z}/157\mathbf Z$ is naturally the direct product of a cyclic group $H$ of order 12 and a cyclic group $K$ of order 13. But $H$ consists of the 13th powers in $k^*$, and $K$ of the 12th powers, so $x^{12}$ is in $K$, and $y^{13}$ is in $H$. But $H\cap K = \{1\}$, so we must have $x^{12}= y^{13}=1$ mod 157.

In particular, both $x^{13}-x$ and $y^{14}-y$ are zero mod 157 and you’re done.

Answer 2

$\!\bmod \color{#90f}{p\!=\!n^2\!+\!n\!+\!1}\!:\ \left[y^{\color{#0a0}{n+1}}\equiv x^{\color{#c00}n}\right]^{\color{#c00}{n+1}}\!\!\Rightarrow y^{\color{#0a0}{n+1}}\equiv \color{#c00}1\Rightarrow x^{n}\,\equiv\, 1\, $ by lil Fermat, for $\,x,y\not\equiv 0,\,$ by

$\!\!\begin{align} \text{ expt arithmetic: }\bmod p\!-\!1\!:\ &\color{#c00}{n(n\!+\!1)} = p\!-\!1\equiv \color{#c00}0,\ \ \ {\rm so}\ \ x^{\color{#c00}{n(n+1)}}\equiv x^{\color{#c00}0}\equiv \color{#c00}1\\[.2em] \underset{{\rm add}\ \color{#0a0}{n+1}}\Longrightarrow\, &(\color{#0a0}{n\!+\!1})(\color{#c00}{n\!+\!1})\equiv \color{#0a0}{n\!+\!1},\,\ {\rm so}\ \ y^{(\color{#0a0}{n+1})(\color{#c00}{n+1})}\!\equiv y^{\color{#0a0}{n+1}}\end{align} $

Thus $\ x\cdot x^n - y\cdot y^{n+1}\equiv x\cdot 1 - y\cdot 1\ $ proves OP (for $\color{#90f}{\rm prime}\ p$). More generally:

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$\!\left.\begin{align}{\bf Lemma}\ \bmod m\!:&\ \ y^k\equiv x^n,\ (x,m)\!=\!1\\ \bmod{\phi=\phi(m)}\!:&\ \ \color{#c00}{nk}\equiv 0,\ \ \ \color{#0a0}{kk\mid k}\end{align}\right\}\,\Rightarrow\, y^k\equiv 1$

Proof: $\ \ \ \bmod m\!: \ y^{kk}\!\equiv x^{\color{#c00}{nk}}\equiv 1\ $ so $\,y^k\equiv 1,\,$ by $\,\color{#0a0}{kk\mid k}\pmod{\!\phi},\,$ where we applied the Congruence Power Rule, and Euler's totient $(\phi)$ theorem, and mod order reduction.

Cor. $ $ Lemma holds if $\,\phi \! =\! \bar nk,\ (k,\bar n)\!=\!1,\ \bar n\mid n,\,$ e.g. $\rm\color{#90f}{prime}$ $\,\color{#90f}{m \!=\! kn\!+\!1,\ k \!=\! n\!+\!1}\,$ (OP).
Proof $\ \phi^{\phantom{|^{|^|}}}\!\!\!\!=\bar nk\mid \color{#c00}{nk}.\,$ $(k,\bar n)\!=\!1\Rightarrow jk = 1 \!+\! i\:\!\bar n$ $ \underset{\times\,k}\Rightarrow jk^2\! = k\! +\! i\phi\Rightarrow \color{#0a0}{k^2\mid k}\pmod{\!\phi}$

Answer 3

Leave out the trivial case when $157|x,y$. Take the primitive root $g$ of $\mod 157$. Therefore we know that if $x^{12}\equiv g^k\mod 157$, $k$ is a multiple of $12$. If $y^{13}\equiv g^k\mod 157$, $k$ is a multiple of $13$. Since $x^{12} \equiv y^{13} \equiv g^k\mod 157$, $k$ is a multiple of $12$ and a multiple of $13$, which is a multiple of $156$, which is $0$. Therefore, $x^{12} \equiv y^{13} \equiv 1\mod 157$, and $x^{13}-y^{14}-x+y\equiv x-y-x+y\equiv 0\mod 157$.

If you need a solution WITHOUT primitive root, we can do it like this: we also leave out the trivial case when $157|x,y$. Then, wun the following code in Python:

set((x**12)%157 for x in range(1,156))
Out[1]: {1, 14, 16, 39, 46, 67, 75, 93, 99, 101, 108, 130, 153}

set((y**13)%157 for y in range(1,156))
Out[2]: {1, 12, 13, 22, 28, 50, 107, 129, 135, 144, 145, 156}

Here you can see the intersection of set of all the $x^{12}$s and the set of all $y^{13}$'s is only $\{1\}$. This also yields $x^{12} \equiv y^{13} \equiv 1\mod 157$