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Problem Of The Week #462 April 5th 2021

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
No one answered last two week's POTW. (Sadface) However, you can find the suggested solution as follows:
Denote $f(n)=2n^3+4n^2-3n+12$. The following table shows the remainders of $n^2,\,n^3,\,n^5$ and $f(n)$ upon division by 11:
\begin{equation*}
\begin{array}{c|c|c|c|c}
n & 0 & 1 & 2 &3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
n^2 & 0 & 1 & 4 & 9 & 9 & 3 & 3 & 5 & 9 & 1 & 1 \\
n^3 & 0 & 1 & 8 & 5 & 9 & 4 & 7 & 2 & 6 & 3 & 10 \\
n^5 & 0 & 1 & 10 & 1 & 1 & 1 & 10 & 10 & 10 & 1 & 10 \\
f(n) & 1 & 4 & 5 & 5 & 5 & 6 & 9 & 4 & 3 & 7 & 6 \\
\end{array}
\end{equation*}

As one can see from the table, the only remainders upon division by 11 that the fifth power of an arbitrary integer $n$ can give are 0, 1, and 10. On the other hand, integers of the form $f(n)$ give only remainders 1, 3, 4, 5, 6, 7, and 9 upon division by 11, whereby the remainder is 1 only if $n$ is divisible by 11. Consequently, $f(p)$ can be the fifth power of an integer only if $p$ is divisible by 11. As $p$ is prime, the only possibility is $p=11$. And indeed, $f(11)=2\cdot 11^3+4\cdot 11^2-3\cdot 11+12=3125=5^5$.
 
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