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3
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Find the remainder when the $2006! + \dfrac{4012!}{2006!}$ is divided by $4013$
$$2006!+\frac{4012!}{2006!}=x \pmod{4013}$$
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!=...
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3
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0
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If 13 does not divide m, then prove that $m^4+8$ is not a cube of an integer [closed]
My question is how can we prove that $m^4 + 8$ not a cube of an integer if
$m$ can not be divided by 13.
What I have done so far:
By Fermat’s Little Theorem:
\begin{align}
m^{p-1} &\equiv 1 \...
- 111
2
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Proof involving modular and primes
My Question Reads:
If $a, b$ are integers such that $a \equiv b \pmod p$ for every positive prime $p$, prove that $a = b$.
I started by stating $a, b \in \mathbb Z$.
From there I have said without ...
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