A Counterexample to the statement, If a≡b±odm and c≡d±odm, where a,b,c,d, and m are integers with c and d positive and m≥2, then ac≡bd±odm.

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 a≡b±odm We get, 2≡2±od5, thus this congruence is satisfied. Next let us plug these values in c≡d±odm We get, 4≡9±od5, observe that 9=4+1×5, thus 4≡4±od5, again this congruence is also satisfied. Lastly we plug these values into ac≡bd±odm We get, 24≡29±od5 16≡512±od5, observe that 16=1+3×5 and 512=2+102×5, thus 1±od5≡2±od5, 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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glamrockqueen7

Answered question

2021-02-27

A Counterexample to the statement, If

a \equiv b \pm o d m

and

c \equiv d \pm o d 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} \pm o d m

Answer & Explanation

Ayesha Gomez

Skilled2021-03-01Added 104 answers

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

a \equiv b \pm o d m

We get,

2 \equiv 2 \pm o d 5

, thus this congruence is satisfied.
Next let us plug these values in

c \equiv d \pm o d m

We get,

4 \equiv 9 \pm o d 5

, observe that

9 = 4 + 1 \times 5

, thus

4 \equiv 4 \pm o d 5

, again this congruence is also satisfied.
Lastly we plug these values into

a^{c} \equiv b^{d} \pm o d m

We get,

2^{4} \equiv 2^{9} \pm o d 5

16 \equiv 512 \pm o d 5

, observe that

16 = 1 + 3 \times 5

and

512 = 2 + 102 \times 5

, thus

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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