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If a is congruent to b (mod3n) prove that a^3 is congruent to b^3 (mod9n)
Solution
a is congruent to b mod(3n)
So, 3n divides (b-a)
To Prove: a 3
is congruent to b 3
mod(9n)
So, we need to prove 9n divides (b 3
- a 3
)
Now, (b 3
- a 3
) = (b - a)( b 2
+ ab + a 2
) ------------(1)
since in R.H.S 3n divides (b-a),
So we only need to show that 3 divides ( b 2 + ab + a 2 )
We can write (b-a) = 3nq , for some integer q
or b = 3nq + a
So, ( b 2
+ ab + a 2
) = [ (3nq+a) 2
+ a(3nq+a) + a 2
]
= 9n 2
q 2
+ 6nq + 3nq 2
+ 3a 2
= 3 (3n 2
q 2
+ 2nq + nq 2
+ a 2
)
So, 3 divides ( b 2
+ ab + a 2
)
Hence proved
:)
If a is congruent to b (mod3n) prove that a^3 is congruent to b^3 (mod.docx

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If a is congruent to b (mod3n) prove that a^3 is congruent to b^3 (mod.docx

  • 1. If a is congruent to b (mod3n) prove that a^3 is congruent to b^3 (mod9n) Solution a is congruent to b mod(3n) So, 3n divides (b-a) To Prove: a 3 is congruent to b 3 mod(9n) So, we need to prove 9n divides (b 3 - a 3 ) Now, (b 3 - a 3 ) = (b - a)( b 2 + ab + a 2 ) ------------(1) since in R.H.S 3n divides (b-a), So we only need to show that 3 divides ( b 2 + ab + a 2 ) We can write (b-a) = 3nq , for some integer q or b = 3nq + a So, ( b 2 + ab + a 2 ) = [ (3nq+a) 2 + a(3nq+a) + a 2 ] = 9n 2 q 2 + 6nq + 3nq 2 + 3a 2 = 3 (3n 2 q 2 + 2nq + nq 2 + a 2 ) So, 3 divides ( b 2 + ab + a 2 ) Hence proved :)