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If a and b are positive integers, is a^4-b^4 divisible by 4?

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If a and b are positive integers, is a^4-b^4 divisible by 4? [#permalink]

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If a and b are positive integers, is \(a^4 - b^4\) divisible by 4?

1) a+b is divisible by 4

2) The remainder is 2 when \(a^2 + b^2\) is divided by 4
[Reveal] Spoiler: OA

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Last edited by Bunuel on 22 Mar 2017, 23:23, edited 1 time in total.
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Re: If a and b are positive integers, is a^4-b^4 divisible by 4? [#permalink]

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ziyuen wrote:
If a and b are positive integers, is \(a^4 - b^4\) divisible by 4?

1) a+b is divisible by 4

2) The remainder is 2 when \(a^2 + b^2\) is divided by 4


Hi

Very good queston.

\(a^4 - b^4 = (a + b)(a - b)(a^2 - b^2)\)

(1) a+b is divisible by 4

Straightforward and sufficient.

(2) The remainder is 2 when \(a^2 + b^2\) is divided by 4

This is a bit tricky.

We need to be aware of the properties of remainders.

If a=r (mod c), has a remainder \(r\) when divided by \(c\), then \(a^2\) will have remainder \(r^2\) when divided by c, given c > r^2.

In our case when c=4 we can have remainders: 0, 1, 2, 3.

a = r (mod 4), b = q (mod 4) ------> \(a^2 = r^2\) (mod 4), \(b^2 = q^2\) (mod 4)

We are given \(r^2 + q^2 = 2\). There is only one possibility \(1^2 + 1^2 = 1 + 1 = 2\) (We can't get \(0 + 2\) or \(2 + 0\) because that means that \(a\) has a remainder \(\sqrt{2}\) when divided by \(4\) and this is impossible).

Hence a^2 - b^2 = 1 - 1 = 0 is completely divisible by \(4\). Sufficient.

Answer D.

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If a and b are positive integers, is a^4-b^4 divisible by 4? [#permalink]

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\(a^4−b^4=(a+b)(a−b)(a^2+b^2)\)

1. \(a+b\) is divisible by 4

Straightforward and sufficient.


2. Let \(a^2+b^2 = 4x+2\) (since it gives 2 as remainder when divided by 4)

So, \(a^{4}−b^{4} = (4x+2)(a+b)(a-b)\) which breaks down to ..\(2(2x+1)(a+b)(a-b)\)

Now, any number which has product of two even numbers will be divisible by 4.
We already have 2. So we need to know if we have another even number or not.

So, 2x+1 .. always odd
(a+b) and (a-b) both will be even if a & b both are odd or both are even.

Since \(a^2+b^2\) gives a remainder of 2 when divided by 4, it means the sum sum is even.

Now, the sum will be even only if \(a^2\) & \(b^2\) both are even or both are odd.

That means, a & b both are even or both are odd (since square of odd numbers are odd and even numbers are even).

So, sufficient.

Hence, the answer is D.

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Re: If a and b are positive integers, is a^4-b^4 divisible by 4? [#permalink]

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New post 19 Nov 2017, 10:29
\((a^4 - b^4\)) % 4 = 0?

\((a^2 + b ^2)(a - b)(a+b)\) % 4 = 0?

Statement 1:
(a + b) is divisible by 4 => (a + b) % 4 = 0 => \((a^2 + b ^2)(a - b)(a+b)\) is divisible by 4

Sufficient

Statement 2:
\((a^2 + b^2)\) % 4 = 2, Let \((a^2 + b^2)\) = 4x + 2 (where x is any integer)

To Note, since remainder is 2, \((a^2 + b^2)\) must be even
=> either \(a^2 hence (a) and b^2 hence(b) both are Odd\)
\(or\)
\(a^2 hence (a) and b^2 hence(b) both are Even\)

Substituting \((a^2 + b^2) = 4x + 2\) in \((a^2 + b ^2)(a - b)(a+b)\)
=> \((4x + 2)(a^2 - b^2)\)
=> \(4x(a^2-b^2) + 2 (a^2 - b^2)\)

First term, 4x(a^2-b^2) is clearly divisble by 4
Second term, \(2 (a^2 - b^2)\) , since we know a, b are either both odd or even, so that \((a^2 - b^2)\)is even => \(2 * even\) => clearly divisible by 4

Sufficient

Answer (D)

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Re: If a and b are positive integers, is a^4-b^4 divisible by 4?   [#permalink] 19 Nov 2017, 10:29
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