System of congruence equations

Question

I have a system of congruence eqs

$$ \begin{cases} x \equiv 14 \pmod{98} \\ x \equiv 1 \pmod{28} \end{cases} $$

I have calculated $\text{gcd}(98,28) = 14$.

I can from the congruence eqs get $x = 14+98k$ and $x = 1+28m$.

I equate these

$$ 14+98k = 1+28m \Leftrightarrow 28m - 98k = 13 $$

I know that $\text{gcd}(98,28) = 14$ is not divisible by 13 and therefore the system has no solutions.. Is this correct?

Answer 1

It is correct. If you look at $28m-98k=2(14m-49k)=13$, you immediately have the left hand side being even whereas the right hand side is odd, contradiction.

Answer 2

we get by definition $$x=14+98k_1$$ and $$x=1+28k_2$$ with $k_1,k_2$ are integers, thus we get $$98k_1-28k_2=-13$$ this equation has no solutions, why?

Answer 3

You are right, but this is faster:

$x\equiv 14\pmod{98}$ implies that $x$ is a multiple of $7$, but $x\equiv 1\pmod{28}$ implies that it isn't.

Answer 4

As $(28,98)=14$

$$x\equiv14\pmod{98}\equiv14\pmod{14}\equiv0$$

and $$x\equiv1\pmod{28}\equiv1\pmod{14}$$ which is impossible