nkhl.goyal wrote:
For positive integers m and n, is m^2 - n^2 divisible by 3?
(1) m - n is divisible by 3
(2) When m^2 + n^2 is divided by 3, the remainder is 2
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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
The questions asks if \(m^2-n^2 = (m-n)(m+n)\) is divisible by \(3\).
Since \(m-n\) is divisible by \(3\) from condition 1), \(m^2-n^2 = (m-n)(m+n)\) is divisible by \(3\) and condition 1) is sufficient.
Condition 2)
If \(k\) is not divisible by \(3\), \(k^2\) has a remainder \(1\) when it is divided by \(3\) for the followings.
If \(k = 3a + 1\), then \(k^2=(3a+1)^2 = 9a^2+6a+1 = 3(3a^2+2a)+1\).
If \(k = 3a + 2\), then \(k^2=(3a+2)^2 = 9a^2+12a+4 = 3(3a^2+4a+1)+1\).
Thus, the condition "\(m^2+n^2\) has remainder \(2\) when it is divided by \(3\)" means that \(m\) and \(n\) are not divisible by \(3\).
Then we have four cases.
Case 1: \(m=3b+1, n=3c+1\)
\((m-n)(m+n)=(3b+1-3c-1)(3b+1+3c+1)=(3b-3c)(3b+3c+2)=3(b-c)(3b+3c+2)\) is divisible by \(3\).
Case 2: \(m=3b+1, n=3c+2\)
\((m-n)(m+n)=(3b+1-3c-2)(3b+1+3c+2)=(3b-3c-1)(3b+3c+3)=3(3b-3c-1)(b+c+1)\) is divisible by \(3\).
Case 3: \(m=3b+2, n=3c+1\)
\((m-n)(m+n)=(3b+2-3c-1)(3b+2+3c+1)=(3b-3c+1)(3b+3c+3)=3(3b-3c+1)(b+c+1)\) is divisible by \(3\).
Case 4: \(m=3b+2, n=3c+2\)
\((m-n)(m+n)=(3b+2-3c-2)(3b+2+3c+2)=(3b-3c)(3b+3c+3)=3(b-c)(3b+3c+4)\) is divisible by \(3\).
Since condition 2) yields a unique solution, it is sufficient.
Therefore, D is the answer.