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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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New post Updated on: 23 Mar 2017, 00:23
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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

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Originally posted by hazelnut on 23 Mar 2017, 00:22.
Last edited by Bunuel on 23 Mar 2017, 00:23, edited 1 time in total.
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If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

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New post 24 Mar 2017, 06:24
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4
\(a^4−b^4=(a+b)(a−b)(a^2+b^2)\)

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 23 Mar 2017, 02:50
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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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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, 11:29
1
\((a^4 - b^4\)) % 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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If a and b are positive integers, is a^4-b^4 divisible by 4?  [#permalink]

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New post 28 Mar 2018, 05:47
vitaliyGMAT wrote:
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.


vitaliyGMAT
Although you have made a good analysis, Your initial expression \(a^4 - b^4 = (a + b)(a - b)(a^2 - b^2)\) is not correct.
Correct form of expression: \(a^4 - b^4 = (a + b)(a - b)(a^2 + b^2)\)
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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 26 Apr 2018, 19:41
itisSheldon wrote:
vitaliyGMAT wrote:
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.


vitaliyGMAT
Although you have made a good analysis, Your initial expression \(a^4 - b^4 = (a + b)(a - b)(a^2 - b^2)\) is not correct.
Correct form of expression: \(a^4 - b^4 = (a + b)(a - b)(a^2 + b^2)\)


Can anyone please explain why the remainder = 2, a^2 + b^2 both are odd or both are even?
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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 26 Apr 2018, 21:58
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hazelnut 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,
Using the formula \(a^{2} - b^{2} = (a+b)(a-b)\), we can simplify the given expression as follows:

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

(1) a+b is divisible by 4

Sufficient.

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

=> \(a^{2} + b^{2} = 4k + 2 = 2(2k + 1)\) (where k is an integer) --- (1)

=> \(a^{2} = 4k + 2 - b^{2}\)

\(a^{2} - b^{2}\) can be written as \(4k + 2 - b^{2} - b^{2} = 4k + 2 - 2b^{2} = 2(2k + 1 - b^{2})\) --- (2)

By using equation (1) and (2) we can write \(a^4 - b^4\) as follows:

\(a^4 - b^4 = (a^{2} + b^{2})(a^{2} - b^{2}) = 2(2k + 1) \times 2(2k + 1 - b^{2}) = 4 (2k + 1)(2k + 1 - b^{2})\)

Hence, divisible by 4. Sufficient.

Thanks.
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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 27 Apr 2018, 14:09
Hi akara2500
As per Statement 2 :
When a^2 + b^2 is divisible by 4, remainder =2.
Hence a^2+ b^2 = 4k+2, thats is even.
Hence either both a,b are even or both a,b are odd.
If both are even, a =2p, b =2q. Hence a^4- b^4 = 16(p^4-q^4). Hence divisible by 4.
If both are odd, a=(2m+1), b=(2n+1)
Now a^2-b^2=4(m^2-n^2+m-n). Hence divisible by 4. Also a^4-b^4= (a^2-b^2)(a^2+b^2). Hence divisible by 4.
Statement 2 is also sufficient.

Answer D
akara2500 wrote:
itisSheldon wrote:
vitaliyGMAT 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.


vitaliyGMAT
Although you have made a good analysis, Your initial expression \(a^4 - b^4 = (a + b)(a - b)(a^2 - b^2)\) is not correct.
Correct form of expression: \(a^4 - b^4 = (a + b)(a - b)(a^2 + b^2)\)


Can anyone please explain why the remainder = 2, a^2 + b^2 both are odd or both are even?[/quote]
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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 06 May 2018, 21:22
\(a^4−b^4=(a+b)(a−b)(a^2+b^2)\) ---------1

1. 1st statement is very simple and straight forward. And it is sufficient.

2. Let see how we will do for 2nd Statement.

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

where x is the quotient and 2 is the remainder.

Using 2 in 1 we get \(a^4−b^4=(4x+2)(a+b)(a−b)\)

Which breaks down to \(a^4−b^4=2(2x+1)(a+b)(a−b)\)

We have to find out if eq 1 is divisible by 4.
Product of any two even numbers will be divisible by 4.


We have one even number- 2. So we need to find out if any of the expression (2x+1), (a+b) & (a-b) will give us even number?

We know that 2x+1 is always odd. NOTE - it is general representation of ODD numbers.

In what scenario (a+b) & (a-b) will give us odd/even.
1) If a and b is odd. Then both (a+b) and (a-b) will be even.
2) If a and b is even. Then both (a+b) and (a-b) will be even.


But how can we decide if a & b is even or odd?

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

From here we have to backtrack with logic to determine if a & b is odd or even.

Again, sum will be even if both the numbers are odd or both the numbers are even.

There are two scenario both leading to common results:-
1) If the square of a number is odd. then the number itself will be odd.
2) Similarly, if square of the numbers are even then, the numbers itself will be even.

From scenario 1 :- Since sum of expression \(a^2+b^2\) is even, let's assume both \(a^2\) and \(b^2\) are odd. Then it implies both a & b are odd. Also the expression (a+b) & (a-b) will be even.

From scenario 1 :- Since sum of expression \(a^2+b^2\) is even, let's assume both \(a^2\) and \(b^2\) are even. Then it implies both a & b are even. Also the expression (a+b) & (a-b) will be even.

Hence the expression \(a^4−b^4\) is divisible by 4. So, sufficient.


Hence, the answer is D.

Hope this helps!!
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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 06 Oct 2018, 10:25
1

Official Explanation:


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

Statement 1 :

a+b is divisible by 4, a+b = 4k
Since, \(a^4-b^4=(a-b)(a+b)(a^2+b^2)\),
we get, \(a^4-b^4=4k(a-b)(a^2+b^2)\)
Hence it is divisible by 4,
SUFFICIENT


Statement 2 :

\(a^2+b^2\) , when divided by 4 gives 2 as remainder, Hence \(a^2+b^2\) is Even.
It means either both a and b are EVEN or ODD.
When both are Even, a = 2m, b = 2n: \(a^4-b^4 = 4(m^4-n^4)\), divisible by 4
When both are oddm a = 2m +1, b = 2n+1:
\(a^4-b^4=(a-b)(a+b)(a^2+b^2)\)
=\(4(m-n)(m+n+1)(a^2+b^2)\), hence 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]

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New post 06 Oct 2018, 10:29
a4−b4 = (a+b)*(a-b) (a2+b2)

Option A: a+b is divisible by 4
since a and b are integers, it definitely answers that a4-b4 will be divisible by 4

2) The remainder is 2 when a2+b2 is divisible by 4
which means a2+b2=. 4k +2

so it becomes (a+b)(a-b) (4k+2)
can't say that it is divisible by 4


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

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New post 06 Oct 2018, 10:35
a^4 - b^4 = (a-b)(a+b)(a^2-b^2)

St 1 says a+4 div by 4 - sufficient

St 2 says a^2+b^2mod 4 = 2 (remainder 2)
which means a and b are either both odd, or both even
Both cases imply a+b is even Hence product of two even numbers is div by 4

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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 06 Oct 2018, 10:36
a^4-b^4 can be factorized as (a^2-b^2)*(a^2+b^2)... (a-b)*(a+b)*(a^2+b^2)

St.1 says a+b is divisible by 4. Hence sufficient.

St.2 says remainder is 2 when a^2+b^2 is divided by 4.. hence it is divisible by 2. It could have both even or both odd parts adding up to an even total... Hence either ways a+b and a-b will be even. Therefore the whole will be divisible by 4.

St2. Is also sufficient.

Hence option (d) is correct.

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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 06 Oct 2018, 10:53
Answer 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 06 Oct 2018, 10:59
a^4-b^4=(a^2+b^2)(a+b)(a-b)

1) suff
2)Suff
(a^2+b^2) div 4..... rem 2 .....===> (a^2+b^2) is divisible by 2 not by 4
case 1 : so a, b both even .... then (a+b) & (a-b) both div by 2
case 2 : so a, b both odd .... then (a+b) & (a-b) both div by 2

Hence ans D
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Re: If a and b are positive integers, is a^4-b^4 divisible by 4?   [#permalink] 06 Oct 2018, 10:59
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