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$$2006!+\frac{4012!}{2006!}=x \pmod{4013}$$

Answer: $x=1553.$

Solution: $$2006!+4012!/2006!=x\pmod{4013}$$ $$(2006!)^2 -2006!x+4012!=0\pmod{4013} (*)$$ $$4\cdot (2006!)^2-4\cdot 2006!x+4\cdot 4012!=0$$ $$(2\cdot 2006!-x)^2-x^2+4\cdot 40121=0$$ $$(2\cdot 2006!-x)^2=x^2-4\cdot 4012! \pmod{4013}(1)$$ For $x=1553$ we have $$x^2=1553^2=601\cdot 4013-4 \implies 1553^2=-4\pmod{4013}.$$ On the other hands, Since $4013$ is prime number,by Wilson's theorem we have $(4012!+1);$ $4013$ It follows that $$4012!=-1\pmod{4013} (2)$$ $$4\cdot 4012!=-4\pmod{4013}$$ From $(1),(2),(3)$ we get $$(2\cdot 2006!-1553)^2=1553^2-4\cdot 4012!\pmod{4013} =-4-(-4)=0\pmod{4013}$$ $$(2\cdot 2016!-1553)^2=0\pmod{4013}$$ $$2\cdot 2016!-1553=\pmod{4013}$$ The last equality is always true because of $$2\cdot 2016!=2\cdot (1\cdot 2\cdot 3\cdot \ldots \cdot 2016) 1553$$ Thus,we proved that $(1)$ is true for $x=1553$ or $(*)$ is true for $x=1553.$

Therefore $$2006!+ 4012!/2006!=1553\pmod{4013}*$$

I need a method to solve the example without using the answer, please with full proof.

Aaratrick
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user825769
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  • I need a method to solve the example without using the answer, please with full proof. – user825769 Sep 16 '20 at 13:10
  • It is unclear what you are asking. The title is just a "word salad" and the body of your post simply presents a calculation. The only hint as to what you want is in the Comment above. Please use the body of your post to explain the setup and goal of the problem you want help with, and give the context so Readers can respond efficiently. – hardmath Sep 16 '20 at 13:51
  • (2006! +4012! / 2006!) Find the remainder when the number is 4013 – user825769 Sep 16 '20 at 14:02
  • The given solution can be shortened like : To find the value of $x$ in $$2006!+\frac{4012!}{2006!}=x \pmod{4013}$$

    $$ \begin{matrix} 2006!+(-1-2\dots -2006) &=x \pmod{4013}\ 2*2006! &= x \pmod{4013} \end{matrix} $$

    But proceeding further without a calculator seems difficult (1553 is prime)

     – Anindya Prithvi Sep 16 '20 at 14:49
  • Mr. This is not a solution I worked using the answer Do you understand I need to calculate the example without using the answer – user825769 Sep 16 '20 at 15:08
  • Thank you for your feedback – user825769 Sep 16 '20 at 15:10
  • As mentioned above, $2006! = \frac{4012!}{2006!}$ mod $4013$, so by Wilson's theorem $(2006!)^2 = -1$. By trial and error I found that $62^2+13^2 = 4013$. You can then multiply $62$ by the inverse of $13$ $($mod $4013)$ to get a square root of $-1$, thus you get both of them. The trouble is determining which square root of $-1$ $2006!$ turns out to be! – Al Suarizmi Sep 16 '20 at 16:17
  • Step by step solution please – user825769 Sep 16 '20 at 16:35
  • @AlS For more on computing $((p-1)/2)! \equiv \pm1 \pmod p,$ for a prime $,p\equiv 1\pmod{4},$ see here and here. – Bill Dubuque Sep 16 '20 at 23:30
  • 2 • 2006! = X (mod 4013) Please prove how we calculate this. – user825769 Sep 17 '20 at 12:25

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