Given $n^{12} - n^8 - n^4 + 1$ it's easy to factorize it: $(n-1)^2(n+1)^2(n^2+1)^2(n^4+1)$. It's stated than this should be divisible by $2^k$ for any odd $n$. So I think that such maximum $k$ can be found by putting the least possible odd $n$ in the expression (because we can easily find such a $k$, that for a bigger $n$ the expression above is divisible by $2^k$, but for a smaller $n$ it isn't). So the least possible $n=3$ (for $n=1$ expression is $0$) and we have $4 \cdot 16 \cdot 100 \cdot 82 = 25 \cdot 41 \cdot 2^9$, so the $k$ we are looking for is $9$. I am rather weak in number theory so I hope the get some feedback if my solution is correct.
Find maximum natural number k such that for any odd n $n^{12} - n^8 - n^4 + 1$ is divisible by $2^k$
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I added an answer showing how to derive it using basic results that work far more generally (double root test and LTE = lifting the exponent). Both are important to know well for number theory and algebra. – Bill Dubuque Mar 18 '21 at 14:55
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First, we start by factorising the expression just like you did:
$$(n-1)^2(n+1)^2(n^2+1)^2(n^4+1)$$
And we know, since $n$ is odd, each of the above terms is even: $(n-1)$, $(n+1)$ are even. $n^2$ and $n^4$ are also odd, so $(n^2 +1)$ and $(n^4+1)$ are even too. This means that we have 7 even factors.
BUT, there's one more important thing here, the two even numbers before and after an odd number, one of them has to be a multiple of 4. And this means that either $(n+1)$ or $(n-1)$ is a multiple of 4, not just 2, and because both terms are squared, we have to add 2 to our original result, 7, which makes our final result 9.
Anas Khaled
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It's $\,(\color{#0a0}{n^4\!+\!1})(\color{#c00}{n^4\!-\!1})^2$ so $\,n\,$ odd $\,\Rightarrow\begin{align}&\ \,2\mid \color{#0a0}{n^4\!+\!1}\\ &2^4\mid \color{#c00}{n^4\!-\!1}\end{align}\,$ since $\,\begin{align} n&\equiv \pm1\ \ \pmod{\!4}\\\Rightarrow \color{#c00}{n^4}&\equiv \color{#c00}{(\pm1)^4}\!\!\!\!\pmod{\!4^2}\end{align}$
Bill Dubuque
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This double-root test method (or LTE method below) work more generally than other answers. $$\large n\equiv \pm1\pmod{!2^2}\Rightarrow,n^2\equiv 1\pmod{!2^3}\Rightarrow n^4\equiv 1\pmod{!2^4}\qquad$$ – Bill Dubuque Mar 18 '21 at 14:52
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With $n:=2m+1$, the polynomial is
$$(2m)^2(2m+2)^2(4m^2+4m+2)^2(16m^4+32m^3+24m^2+8m+2) \\=2^7m^2(m+1)^2(2m^2+2m+1)(8m^4+16m^3+12m^2+4m+1).$$
The last two factors are always odd and the remaining ones form at least a multiple of $2^9$.
More precisely, if the multiplicity of the factor $2$ in the prime factorization of $\dfrac{n-1}2$ is $p>0$, then $k=2p+7$, otherwise $k=2q+7$ where $q$ is the multiplicity of $2$ in $\dfrac{n+1}2$.
E.g., with $n=15$, $p=0,q=3$ and $k=13$. Indeed, $15^{12}-15^8-15^4+1=2^{13}\,15837863153$, and with $n=17$, $p=3$ and $17^{12}-17^8-17^4+1=2^{13}\,71120027025$.
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The answer $k=9$ is correct but I don't think your reasoning is good. For example for expression $(n+1)^2(n^2+1)^2(n^4+1)$ this doesn't work because for $n=3$ the result is divided by $2^7$ but for $n=5$ it is divided only by $2^5$. I think the way to solve this is too look at expressions $n-1$, $n+1$, $n^2+1$, $n^4+1$ and try to find what's the least amount of 2's that could be dividing each of them.
szymji
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For $n=7$ the original expression would be divisible by $2^{11}$, we are considering the lowest k valid for any odd $n$. In your edited expression $k$ would be 5, each factor has to be even, so you can take out a 2 from each one. – Anas Khaled Feb 01 '24 at 06:13
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$$(2r+1)^2+1\equiv8\cdot\dfrac{r(r+1)}2+2=8p+1\text{(say)}\equiv2\pmod8$$
$$(2r+1)^4+1=(8p+1)^2+1\equiv2\pmod{16}$$
Now $n^2-1=(2r+1)^2-1=8\cdot\dfrac{r(r+1)}2$
$\implies k\ge2\cdot3+2\cdot1+1$
lab bhattacharjee
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