I already know that $(n^3 - n)$ is divisible by 6. I have a hunch that the second parenthesis of 5 and 3 is divisible by $5^4 + 3^2.$ I jus need a hint and clarity on how to proceed further. Plz help me.
Given that $(n^3 - n)(5^{8n+4} + 3^{4n+2})-2017$ gives the same remainder(R) when divided by 3804 for all integers n>=2 then what is the value of R?
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number-theory
elementary-number-theory
2 Answers
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We are given that $(n^3-n)(5^{8n+4}+3^{4n+2})-2017\equiv R$ mod $3084$ $\forall n\geq 2$. Now, in particular for $n=3084$, we see that
$(n^3-n)(5^{8n+4}+3^{4n+2})=n(n^2-1)(5^{8n+4}+3^{4n+2})\equiv0$ mod $3084$ and $2017\equiv -1067$ mod $3084$, hence $R\equiv 1067$ mod $3084$.
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0Thank a lot......Well how can u all solve these probs so easily? – 2017-01-04
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0Well it's 3804 and not 3084.... Thus the answer is 1787 – 2017-01-04
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0But then......U told R is congruent to 1067 modulo 3084...... But the question is R is how much? So does this satisfy the solution? – 2017-01-04
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0Well then take $n=3804$ and hence $R\equiv 1787$ mod $3804$. Yes it answers your question as then you can take $R=1787$. – 2017-01-04
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Hint: You can plug in any value $n\geq 2$ you want to find the answer. What's a value of $n$ that makes the big nasty expression simple mod $3084$?
A bigger hint is hidden below.
Plug in $n=3084$.