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# A Counterexample to the statement, If a\equiv b\pmod m and c\equiv d\pmod m, where a,b,c,d, and m are integers with c and d positive and m \geq 2, then a^{c}\equiv b^{d}\pmod m.

Congruence
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asked 2021-02-27
A Counterexample to the statement, If $$\displaystyle{a}\equiv{b}\pm{o}{d}{m}$$ and $$\displaystyle{c}\equiv{d}\pm{o}{d}{m}$$, where a,b,c,d, and m are integers with c and d positive and $$\displaystyle{m}\geq{2}$$, then $$\displaystyle{a}^{{{c}}}\equiv{b}^{{{d}}}\pm{o}{d}{m}$$.

## Answers (1)

2021-03-01
Here we need to just find one counter example that contradicts the validity of the given statement.
Suppose we set values a = 2, b=2, c= 4, d=9, and m=5 in
$$\displaystyle{a}\equiv{b}\pm{o}{d}{m}$$
We get,
$$\displaystyle{2}\equiv{2}\pm{o}{d}{5}$$, thus this congruence is satisfied.
Next let us plug these values in $$\displaystyle{c}\equiv{d}\pm{o}{d}{m}$$
We get,
$$\displaystyle{4}\equiv{9}\pm{o}{d}{5}$$, observe that $$\displaystyle{9}={4}+{1}\times{5}$$, thus
$$\displaystyle{4}\equiv{4}\pm{o}{d}{5}$$, again this congruence is also satisfied.
Lastly we plug these values into $$\displaystyle{a}^{{{c}}}\equiv{b}^{{{d}}}\pm{o}{d}{m}$$
We get,
$$\displaystyle{2}^{{{4}}}\equiv{2}^{{{9}}}\pm{o}{d}{5}$$
$$\displaystyle{16}\equiv{512}\pm{o}{d}{5}$$, observe that
$$\displaystyle{16}={1}+{3}\times{5}$$ and $$\displaystyle{512}={2}+{102}\times{5}$$, thus
$$\displaystyle{1}\pm{o}{d}{5}\equiv{2}\pm{o}{d}{5}$$, which is incorrect
Thus a counterexample could be the set of values
a=2, b=2, c=4, d=9, and m =5

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