Help solving congruence equation (modulo).

Question

If $x \in \mathbb{Z}$ and we have

1. $14x +3 \equiv 5x - 6 \pmod 8$
2. $(5x+1)^2 + 2 \equiv 3 \pmod 6$

How do I find the solution set for $x$?

Answer 1

From the first equation we get that for some $\alpha \in \mathbb{Z}$ $$\begin{align} 14x +3 \equiv 5x - 6 \pmod 8 &\iff 14x +3 = 8\alpha + 5x-6\\ &\iff 9x +9 = 8\alpha\\ &\iff 9(x +1) = 8\alpha\\ &\iff x+1 \equiv 0 \pmod 2\\ &\iff x \equiv 1 \pmod 2 \end{align}$$ and from the second that for some $\beta \in \mathbb{Z}$ $$\begin{align} (5x+1)^2 + 2 \equiv 3 \pmod 6 &\iff (5x+1)^2 + 2 = 6\beta + 3\\ &\iff (5x+1)^2 - 1 = 6\beta\\ &\iff (5x+1)^2 \equiv 1 \pmod 2\\ &\iff 5x+1 \equiv 1 \pmod 2\\ &\iff x \equiv 0 \pmod 2 \end{align}$$ thus we get $$x \equiv 0 \pmod 2 \wedge x \equiv 1 \pmod 2$$ but that's an impossible condition to sussist, so there are no solution for$x \in \mathbb{Z}$.

Answer 2

In fact your system has no solution:

From the first equation we have: $$ 9x \equiv -9 \mod{8} \; \; \; \Rightarrow x\equiv -1 \mod{8}$$ so we may write $$x=8k-1 \text{ with } k \in \mathbb{Z}$$ Substituting in the second equation: $$(5x+1)^2+2= (40k-4)^2+2 \equiv (4k-4 )^2 +2 \equiv 4k(k+1) \mod{6}$$ however if we consider $ \mod{6}$:

if $k\equiv 0 \mod{6}$ then $0(0+1) \not \equiv 3$

if $k\equiv 1 \mod{6}$ then $ 4(1+1)= 8 \equiv 2 \not \equiv 3 $

if $k\equiv 2 \mod{6}$ then $ 8(2+1)=24 \equiv 0 \not \equiv 3$

if $k\equiv 3 \mod{6}$ then $ 12(3+1)\equiv 0 \not \equiv 3$

if $k\equiv 4 \mod{6}$ then $ 16(4+1) \equiv 4\times 5 \equiv 20 \equiv 2 \not \equiv 3$

if $k\equiv 5 \mod{6}$ then $ 20 ( 5+1) \equiv 0 \not \equiv 3$

Hence the above system has no solution.

Answer 3

$$14x +3 \equiv 5x - 6 \pmod 8\implies 2\mid 14x+3-(5x-6)=2(4x+4)+x+1$$

$$(5x+1)^2+2\equiv 3\pmod{6}\implies 2\mid (5x+1)^2+2-3=2\left(12x^2+5x\right)+x^2$$

The first one gives $x$ is odd, while the second one gives $x$ is even. No solutions.