I have a problem regarding the understanding of modulus.
So someone proved $5a + 2b \equiv 0\pmod 7$ is symmetric using the modulo circle, and my brain cant comprehend how he meant that. I'm talking about circles like this:
I know there is a other way to prove it with $7 \mid 5a + 2b$
On
The relation is defined on $\Bbb Z$ like this: $a \sim b$ if and only if $5a + 2b \equiv 0 \pmod 7$. On the assumption this holds, we wish to show $b \sim a$, i.e. $5b+2a \equiv 0 \pmod 7$.
Using the definition of modular congruence, then, $\exists k \in \Bbb Z$ such that
$$\frac{5a+2b-0}{7} = k \iff 5a + 2b = 7k$$
Subtract $7a$ and add $7b$ to both sides. Then you have
$$-2a - 5b = 7(k - a + b)$$
Now multiply both sides by $-1$:
$$2a + 5b = 7(-k+a-b)$$
Thus, there an $\ell \in \Bbb Z$ (namely $\ell = -k+a-b$) such that
$$\frac{5b+2a}{7} = \ell$$
ensuring that $5b + 2a \equiv 0 \pmod 7$, i.e. $b \sim a$. Thus, symmetry.
Now, with this in mind, consider the "circle" for equivalence modulo $7$:
If $5a + 2b \equiv 0 \pmod 7$, then the sum $5a+2b$ is in that topmost notch. Moreover, adding multiples of seven to it will make $5a+2b$ cycle back to that same place. So what we do, then, is add $-7a-7b= 7(-a-b) \in 7 \Bbb Z$ to it. This gives us $-2a-5b \equiv 0 \pmod 7$. However, in this notch of the diagram, if $x$ is in it, so is $-x$. (That is, $x \equiv -x \pmod 7$. I leave it to you to justify why.) Thus, $-(-2a-5b)= 2a +5b \equiv 0 \pmod 7$ as desired.
...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.
modular arithmetic is basically arithmetic done using linear polynomials:$$y\equiv c\bmod m\iff y=mx+c$$ for some integer values of x,m,c, and y. So it's basically saying, in this case that: $$5a+2b=7x$$ if we allow $a=b=x=n$ where n is integer, then we have a solution. Edit more to the point $5=7+(-2)$ so if the original has an answer so will $-2a+2b=7x$