In what condition there is no common solution for congruence system Why there is no x such that $x \equiv 3$(mod 8) and $x \equiv 1$(mod 4)?
Is it because $(4, 8) \neq 1$, which contradict the Chinese Remainder Theorem?
But why?
 A: Note $\, x\equiv 3\pmod 8\,\Rightarrow\ 4\mid 8\mid x-3\,\Rightarrow\, x\equiv 3\pmod 4.\ $ Thus $\, 3\equiv x\equiv  1\pmod 4$ yields the contradiction that $\, 4\mid 3-1.\,$
Similarly if $\, x\equiv a\pmod m,\ x\equiv b\pmod n\,$ then $\, a\equiv x\equiv b\pmod d\,$ for $\,d =\gcd(m,n),\ $ so $\,d\mid a-b\,$ is a necessary condition for the existence of a solution.
This compatibility condition is also a sufficient condition for the existence of solution, and it extends pairwise to any number of congruences - see this answer for a constructive proof (which depends on the key fact that gcd distributes over lcm).
A: It does not "contradict", just not "comply" with the necessary conditions for CRT to hold. 
For your question, just think of what is asked of $x$ : $x=1+4k$ and $x=3+8k'$, with $k$ and $k'$ integers. It leads to finding an integer which is at the same time odd and even.
This outcome is not possible when the modulus are co-primes, because of the Bezout theorem.
A: If $x \equiv 3 \pmod 8$, then we can write $x = 8n + 3$ for some integer $n$.
Thus $x = 4(2n) + 3 = 4m + 3$ for some integer $m$, and hence $x \equiv 3 \pmod 4$.

Since $4$ and $8$ are not co-prime, the CRT does not apply.
