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QUESTION: Find a polynomial $p(x)$ that simultaneously has both the following properties.

$(i)$ When $p(x)$ is divided by $x^{100}$ the remainder is the constant polynomial $1.$

$(ii)$ When $p(x)$ is divided by $(x − 2)^3$ the remainder is the constant polynomial $2.$


MY APPROACH: I recently came to know about the Chinese Remainder Theorem, and was literally amazed by it's wide variety of applications. Now the questions I dealt with so far were composed of numbers and linear congruences. This is the first time I have come across a question that deals with polynomials. I know there are other ways to solve this problem, but I am particularly interested in solving it with the aid of the Chinese Remainder Theorem. How is that theorem applied to polynomials, like this one?

Moreover I have examined that $x^{100}$ and $(x-2)^3$ are coprime to each other. So it ensures that the theorem can be applied safely.

Pardon me, this question has been asked before. But the solution provided was not complete. It did not show the full process as to how to do the sum. All I need is one answer where the full method to tackle with polynomials using the CRT is done (consider this question, for instance).

A detailed answer will be very much helpful in this regard. Also it will be great if you can suggest me some texts or reference from where I can learn how to apply CRT on polynomials..

Thank you so much.

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ADDED: What I get is that , for some $n$ which will eventually be set to 100, that $$ x^n \left(\frac{n^2 + n}{2} x^2 - (2n^2 + 4n)x + (2n^2 + 6n + 4) \right) - 2^{n+2} $$ is divisible by $(x-2)^3$

This can be confirmed by evaluating the function and its first derivative and its second derivative at $x=2$ and showing that each results in zero.

There are tricks for doing this rapidly in contests, perhaps at the previous question you linked.

ORIGINAL: Here are Bezout identities for $x^n$ and $(x-2)^3$ for $n=3,4,5,6,7,8,9,10.$ Seem to be some patterns, especially the denominator needed is always a power of $2.$ $$ \left( x^{3} \right) \left( \frac{ 3 x^{2} - 15 x + 20 }{ 16 } \right) - \left( x^{3} - 6 x^{2} + 12 x - 8 \right) \left( \frac{ 3 x^{2} + 3 x + 2 }{ 16 } \right) = \left( 1 \right) $$ $$ $$ $$ \left( x^{4} \right) \left( \frac{ 5 x^{2} - 24 x + 30 }{ 32 } \right) - \left( x^{3} - 6 x^{2} + 12 x - 8 \right) \left( \frac{ 5 x^{3} + 6 x^{2} + 6 x + 4 }{ 32 } \right) = \left( 1 \right) $$ $$ $$ $$ \left( x^{5} \right) \left( \frac{ 15 x^{2} - 70 x + 84 }{ 128 } \right) - \left( x^{3} - 6 x^{2} + 12 x - 8 \right) \left( \frac{ 15 x^{4} + 20 x^{3} + 24 x^{2} + 24 x + 16 }{ 128 } \right) = \left( 1 \right) $$ $$ $$ $$ \left( x^{6} \right) \left( \frac{ 21 x^{2} - 96 x + 112 }{ 256 } \right) - \left( x^{3} - 6 x^{2} + 12 x - 8 \right) \left( \frac{ 21 x^{5} + 30 x^{4} + 40 x^{3} + 48 x^{2} + 48 x + 32 }{ 256 } \right) = \left( 1 \right) $$ $$ $$ $$ \left( x^{7} \right) \left( \frac{ 14 x^{2} - 63 x + 72 }{ 256 } \right) - \left( x^{3} - 6 x^{2} + 12 x - 8 \right) \left( \frac{ 14 x^{6} + 21 x^{5} + 30 x^{4} + 40 x^{3} + 48 x^{2} + 48 x + 32 }{ 256 } \right) = \left( 1 \right) $$ $$ $$ $$ \left( x^{8} \right) \left( \frac{ 9 x^{2} - 40 x + 45 }{ 256 } \right) - \left( x^{3} - 6 x^{2} + 12 x - 8 \right) \left( \frac{ 9 x^{7} + 14 x^{6} + 21 x^{5} + 30 x^{4} + 40 x^{3} + 48 x^{2} + 48 x + 32 }{ 256 } \right) = \left( 1 \right) $$ $$ $$ $$ \left( x^{9} \right) \left( \frac{ 45 x^{2} - 198 x + 220 }{ 2048 } \right) - \left( x^{3} - 6 x^{2} + 12 x - 8 \right) \left( \frac{ 45 x^{8} + 72 x^{7} + 112 x^{6} + 168 x^{5} + 240 x^{4} + 320 x^{3} + 384 x^{2} + 384 x + 256 }{ 2048 } \right) = \left( 1 \right) $$ $$ $$ $$ \left( x^{10} \right) \left( \frac{ 55 x^{2} - 240 x + 264 }{ 4096 } \right) - \left( x^{3} - 6 x^{2} + 12 x - 8 \right) \left( \frac{ 55 x^{9} + 90 x^{8} + 144 x^{7} + 224 x^{6} + 336 x^{5} + 480 x^{4} + 640 x^{3} + 768 x^{2} + 768 x + 512 }{ 4096 } \right) = \left( 1 \right) $$ $$ $$

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  • $\begingroup$ Woah, this is really some big calculation.. I am yet to know about bezout identities.. thank you for your help though.. :) $\endgroup$ – Stranger Forever May 28 '20 at 19:40
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    $\begingroup$ @StrangerForever given coprime elements $p,q$, a Bezout identity is just a new pair $r,s$ such that $pr+qs = 1,$ arrived at by an Extended Euclidean Algorithm. For polynomials, we allow constant outcome other than $1;$ in any of these, multiply everything by the evident denominator $\endgroup$ – Will Jagy May 28 '20 at 19:46
  • $\begingroup$ Can you suggest me a text where I can read about extended Euclidean algorithm in detail.. I came across the Euclidean division algorithm, and that was pretty interesting.. but I am yet to learn about the extension.. $\endgroup$ – Stranger Forever May 28 '20 at 20:13
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    $\begingroup$ @StrangerForever not sure, first go through the 73 posts I found with both "tags" euclidean-algorithm and polynomials: math.stackexchange.com/questions/tagged/… I have already checked my original function and its first derivative, the pattern is looking very good. I put it at the top of the answer. $\endgroup$ – Will Jagy May 28 '20 at 20:29

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