Question about proof for invariance principle problem

Question

I don't see how they got that $b-c \equiv0$ mod $(3)$ and $a-c \equiv0$ mod $(3)$?

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• Solution

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Answer 1

If $$(a',b',c')= (a+2,b-1,c-1)\implies $$

$$ a'-b' = (a+2)-(b-1) = a-b +3 \equiv_3 a-b $$ and $$a'-c' = (a+2)-(c-1) = a-c +3 \equiv_3 a-c $$ and $$b'-c' = (b-1)-(c-1) = b-c $$

The same holds for other two cases.

Answer 2

Consider for example the transformation $$(a,b,c) \mapsto (a',b',c')=(a+2,b-1,c-1)$$ You can see here that, if you note $I(a,b,c)=a-b\mod 3$, $$I(a',b',c')=a'-b'\mod 3 = (a+2)-(b-1)\mod 3 = a-b+3\mod 3 = a-b\mod 3 = I(a,b,c)$$ So $I$ is an invariant of the transform. You should verify that this is the case whatever triple you chose to transform $(a,b,c)$ to.