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The problem is to simply solve the system of congruences: $$x \equiv 2 \pmod 4,\\ x \equiv 4 \pmod 8.$$ My idea was to write $x=4k+2=8m+4$ for suitable $k,m \in \mathbb{Z}$. Dividing by $2$, $$2k +1 = 4m+2 \Leftrightarrow 2k-1=4m,$$ so that $2k \equiv 1 \pmod 4$. I wanted to reduce it all the way down to get an expression for $k$ and insert this $k$ in $x=4k+2$ and then take the resulting equation modulo the coefficient in front of $k$. But now I have a congruence involving $2k$, how do I get around this?

Hope you can help!

EDIT: Problem solved. Obviously $x \equiv 4 \pmod 8$ implies $x \equiv 0 \pmod 4$, so no solutions. Solved.

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$x \equiv 4$ mod $8$ implies $x \equiv 4$ mod $4$.. –  Cocopuffs Oct 2 '13 at 4:44
Use the chinese remainder theorem but gcd(4,8)=4 so there may or may not be solutions. If there are solutions there will be 4 of them. –  user60887 Oct 2 '13 at 4:46
or you simply observe that $2k-1$ is odd while $4m$ is even... –  N. S. Oct 2 '13 at 4:46
Lol. Obviously. Problem solved. –  Bo Schmidt Oct 2 '13 at 4:46

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