# For the following statement, either prove that they are true or provide a counterexample: Let a, b, c, d, m in Z such that c, d >= 1 and m > 1. If a -= b (mod m) and c -= d (mod m), then a^c -= b^d (mod m)

For the following statement, either prove that they are true or provide a counterexample:
Let a, b, c, d, $m\in Z$ such that c, $d\ge 1$ and m > 1. If $a\equiv b\left(\text{mod}m\right)$ and
$c\equiv d\left(\text{mod}m\right)$, then ${a}^{c}\equiv {b}^{d}\left(\text{mod}m\right)$
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Let a, b, c, d, $m\in Z$ such that c, $d\ge 1$ and m > 1.
If $a\equiv b\left(\text{mod}m\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}c\equiv d\left(\text{mod}m\right)$
${a}^{c}\text{mod}\left(m\right)={\left(a\text{mod}\left(m\right)\right)}^{c}$
${b}^{d}\text{mod}\left(m\right)={\left(b\text{mod}\left(m\right)\right)}^{d}$
But $a\equiv b\left(\text{mod}m\right)$
So we get c = d.
Therefore the statement is true when c =d.
So take c and d are different for a counter example.
Take a = b = 2, c = 16 and d = 6
m=10
Hence $2\equiv 2\left(\text{mod}10\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}16\equiv 6\left(\text{mod}10\right)$
${2}^{16}\mathrm{¬}\equiv {2}^{6}\text{mod}\left(10\right)$
As ${2}^{16}=65536\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{2}^{6}=32$
65536-32 = 65504 which is not divisible by 10.