# If an integer a is such that a-2 is divisible by 3 then a^2-1 is divisible by 3. prove by direct method

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## Solution 1

\$\$a^2-1=(a-1)(a+1)=(a-1)(a-2+3)=(a-1)(a-2)+3(a-1)\$\$

So, \$a^2-1\$ will be divisible by \$3\$ if \$a-1\$ or \$a-2\$ is so

## Solution 2

Hint \$\ \ 3\mid \color{#c00}{a-2}\ \Rightarrow\ 3\mid\!\! \overbrace{a+1}^{\large \ \color{#c00}{a-2}\,+\,3}\!\!\mid a^2-1\$

## Solution 3

We're given

\$\$a-2=3k\iff a= 2+3k\implies a^2-1=(a+1)(a-1)=(3+3k)(a-1)=3(1+k)(a-1)\ldots\$\$

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### Sudeep Acharya

Updated on August 09, 2022

• Sudeep Acharya about 1 year

How to prove that if a is number such that \$a-2\$ is divisible by \$3\$ then \$a^2-1\$ is divisible by \$3\$ using direct method.

I know if \$a = 2\$ then \$a-2 = 0\$ is divisible by \$3\$ and \$2^2-1 = 3\$ is divisible by \$3\$ but how to prove it using direct method.

• Jlamprong almost 10 years
Hint: \$a^2-1=(a-2)(a+2)+3\$
• DonAntonio almost 10 years
If the OP knew even very basic modular arithmetic his question would be completely trivial...
• mathlove almost 10 years
hmm, I agree with you, but I hope this might be the first step for the OP to learn it...
• Sudeep Acharya almost 10 years
yes i can do that thank you.
• lab bhattacharjee almost 10 years
@Downvoter, would you mind sharing the mistake.