# Two congruence systems with a parameter - answer verification

Given two systems of congruences with one common parameter $A$ I need to find all values of $A$ for which both systems have the same solution sets. The systems are: $$x\equiv5\bmod6\\x\equiv A\bmod35$$and$$x\equiv A\bmod10\\x\equiv 14\bmod21$$

I solved it by first finding x in both systems, by CRT and then since the answers were both $\bmod 210$ I equated both solutions with that modulus and solved for $A$.
From the first system I got $x\equiv36A+35\bmod210$, and from the second one $x\equiv21A-70\bmod210$. Equating the $x$'s I got $A\equiv3\bmod6$. Is it correct? And what would I have to do if the moduli weren't equal? Would finding the LCA and multiplying both solutions be correct?

• The solutions of the systems are correct – Peter Aug 30 '17 at 12:42
• For $A=3$, the expressions $36A+35$ and $21A-70$ are not congruent modulo $210$ . $A\equiv 7\mod 14$ seems to be correct. – Peter Aug 30 '17 at 12:48
• @Peter Thanks, I double-checked it again and now I got your answer. – Joald Aug 30 '17 at 12:52
• The case where the moduli are not equal : If they are coprime, you can apply CRT again, have to think about the remaining case – Peter Aug 30 '17 at 12:53

\qquad \begin{align} x\equiv\, \color{#c00}1\!\!\!\pmod 2,\ &\ 2\!\!\!\pmod 3,\ A\!\!\!\pmod 5,\ \color{#0a0}A\!\!\!\pmod 7\\ x\equiv \color{#c00}A\!\!\!\pmod 2,\ &\ 2\!\!\!\pmod 3,\ A\!\!\!\pmod 5,\ \ \color{#0a0}0\!\!\!\pmod 7 \end{align}
By CRT these are equivalent $\iff \color{#c00}A\equiv \color{#c00}1\pmod 2,\ \color{#0a0}A\equiv \color{#0a0}0\pmod 7\,$
Solving we find $\bmod 2\!:\,\ 1\equiv A\equiv 7k\equiv k\,$ so $\, A = 7k = 7(1\!+\!2n) = 7\!+\!14n$