Can "a is congruent to c(mod b)" and "a is congruent to -c(mod b)" both be true at the same time? I think they cannot both be true, but I am wondering the reason why they cannot.
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In general, when $b$ is even, you can take $c = b/2$ and $a=b/2$ and it will be true. – Nov 05 '13 at 00:22
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$a\equiv c \bmod b$ and $a\equiv -c \bmod b$ hold iff $2c \equiv 0 \bmod b$.
So, there are plenty of examples when $b$ is even: just take $a=c=b/2$.
There is also the trivial solution $c\equiv 0 \bmod b$, which works for all $b$. This is the only solution when $b$ is odd.
lhf
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So, what if I make a restriction on b that b is a prime, would it then be impossible for both congruence to be satisfied? – Slavica Nov 05 '13 at 00:26
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I Assume that $b=p$ is prime. Also I assume that p is odd prime. If $p$ is even then you have $1\equiv \pm 1 \mod 2.$ Since $a\equiv c \mod p$ and $a\equiv -c \mod p$ we get $p|\gcd(a+c,a-c)=d.$ But $d|a+c$ and $d|a-c$ thus $d|2a, d|2c$ so $d|2\gcd(a,c).$ If $\gcd(a,c)=1,$ then you get $p|2$ which is a contradiction. So $\gcd(a,c)>1$ so $p|2\gcd(a,c)$ and $\gcd(a,c)=p\cdot r.$ So $a=pr_1,c=pr_2$ (for some integers $r_1,r_2)$ and taking $\mod p$ both $a$ and $c$ zero. That is, if $b=p$ is odd prime, then $a\equiv c\equiv 0 \mod p.$
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