Last visit was: 18 Nov 2025, 14:12 It is currently 18 Nov 2025, 14:12
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
BillyZ
User avatar
Current Student
Joined: 14 Nov 2016
Last visit: 03 May 2025
Posts: 1,143
Own Kudos:
22,214
 [40]
Given Kudos: 926
Location: Malaysia
Concentration: General Management, Strategy
Schools: Sloan '23 (D) Tuck '23 (D) Darden '23 (D) Ross '23 (D) Kellogg '23 (D) Wharton '23 (D) Booth '23 (D) Fuqua '24 (A) Harvard '24 (D) Stanford '24 (D)
GMAT 1: 750 Q51 V40 (Online)
GPA: 3.53
Products:
My Reviews
Schools: Sloan '23 (D) Tuck '23 (D) Darden '23 (D) Ross '23 (D) Kellogg '23 (D) Wharton '23 (D) Booth '23 (D) Fuqua '24 (A) Harvard '24 (D) Stanford '24 (D)
GMAT 1: 750 Q51 V40 (Online)
Posts: 1,143
Kudos: 22,214
 [40]
If a and b are positive integers, is a^4-b^4 divisible by 4? Updated on: 23 Mar 2017, 00:23
Originally posted by BillyZ on 23 Mar 2017, 00:22.
Last edited by Bunuel on 23 Mar 2017, 00:23, edited 1 time in total.
Edited the tags.
5
Kudos
Add Kudos
35
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

85% (hard)

Question Stats:

49% (02:05) correct 51% (01:45) wrong based on 437 sessions

History

Date
Time
Result
Not Attempted Yet
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
ShowHide Answer
Official Answer
D
Most Helpful Reply
avatar
blackfish
avatar
Joined: 02 Feb 2017
Last visit: 08 Feb 2025
Posts: 29
Own Kudos:
29
 [24]
Given Kudos: 9
Location: India
Schools: ISB '20 (D) Foster '21 (A) Fisher '21 (A$)
GMAT 1: 740 Q50 V40
GPA: 2.79
Schools: ISB '20 (D) Foster '21 (A) Fisher '21 (A$)
GMAT 1: 740 Q50 V40
Posts: 29
Kudos: 29
 [24]
If a and b are positive integers, is a^4-b^4 divisible by 4? 24 Mar 2017, 06:24
16
Kudos
Add Kudos
8
Bookmarks
Bookmark this Post
\(a^4−b^4=(a+b)(a−b)(a^2+b^2)\)

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.
User avatar
vitaliyGMAT
User avatar
Joined: 13 Oct 2016
Last visit: 26 Jul 2017
Posts: 297
Own Kudos:
875
 [10]
Given Kudos: 40
GPA: 3.98
Posts: 297
Kudos: 875
 [10]
If a and b are positive integers, is a^4-b^4 divisible by 4? 23 Mar 2017, 02:50
4
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
ziyuen
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
Show more

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.
General Discussion
User avatar
hellosanthosh2k2
User avatar
Joined: 02 Apr 2014
Last visit: 07 Dec 2020
Posts: 361
Own Kudos:
597
 [3]
Given Kudos: 1,227
Location: India
Schools: XLRI"20
GMAT 1: 700 Q50 V34
GPA: 3.5
Schools: XLRI"20
GMAT 1: 700 Q50 V34
Posts: 361
Kudos: 597
 [3]
If a and b are positive integers, is a^4-b^4 divisible by 4? 19 Nov 2017, 11:29
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
\((a^4 - b^4\)) % 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)
User avatar
itisSheldon
User avatar
Joined: 03 Mar 2018
Last visit: 26 Jan 2022
Posts: 161
Own Kudos:
669
Given Kudos: 101
Posts: 161
Kudos: 669
If a and b are positive integers, is a^4-b^4 divisible by 4? 28 Mar 2018, 05:47
Kudos
Add Kudos
Bookmarks
Bookmark this Post
vitaliyGMAT
ziyuen
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.
Show more

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: \(a^4 - b^4 = (a + b)(a - b)(a^2 + b^2)\)
User avatar
akara2500
User avatar
Joined: 29 Mar 2015
Last visit: 19 Dec 2022
Posts: 21
Own Kudos:
3
Given Kudos: 87
GMAT 1: 590 Q44 V23
GMAT 1: 590 Q44 V23
Posts: 21
Kudos: 3
If a and b are positive integers, is a^4-b^4 divisible by 4? 26 Apr 2018, 19:41
Kudos
Add Kudos
Bookmarks
Bookmark this Post
itisSheldon
vitaliyGMAT
ziyuen
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: \(a^4 - b^4 = (a + b)(a - b)(a^2 + b^2)\)
Show more

Can anyone please explain why the remainder = 2, a^2 + b^2 both are odd or both are even?
User avatar
ganand
User avatar
Joined: 17 May 2015
Last visit: 19 Mar 2022
Posts: 198
Own Kudos:
3,677
 [6]
Given Kudos: 85
Posts: 198
Kudos: 3,677
 [6]
If a and b are positive integers, is a^4-b^4 divisible by 4? 26 Apr 2018, 21:58
3
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
hazelnut
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
Show more

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.
User avatar
GMATBusters
User avatar
GMAT Tutor
Joined: 27 Oct 2017
Last visit: 14 Nov 2025
Posts: 1,924
Own Kudos:
6,646
Given Kudos: 241
WE:General Management (Education)
Expert
Expert reply
Posts: 1,924
Kudos: 6,646
If a and b are positive integers, is a^4-b^4 divisible by 4? 27 Apr 2018, 14:09
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
itisSheldon
vitaliyGMAT
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: \(a^4 - b^4 = (a + b)(a - b)(a^2 + b^2)\)
Show more

Can anyone please explain why the remainder = 2, a^2 + b^2 both are odd or both are even?[/quote]
User avatar
SSM700plus
User avatar
Joined: 01 Aug 2017
Last visit: 08 Aug 2022
Posts: 164
Own Kudos:
194
 [1]
Given Kudos: 420
Location: India
Concentration: General Management, Leadership
Schools: ISB '20 (S) IIMA PGPX"20 (S)
GMAT 1: 500 Q47 V15
GPA: 3.4
WE:Information Technology (Computer Software)
Products:
My Reviews
Schools: ISB '20 (S) IIMA PGPX"20 (S)
GMAT 1: 500 Q47 V15
Posts: 164
Kudos: 194
 [1]
If a and b are positive integers, is a^4-b^4 divisible by 4? 06 May 2018, 21:22
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
\(a^4−b^4=(a+b)(a−b)(a^2+b^2)\) ---------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!!
Thanks
User avatar
GMATBusters
User avatar
GMAT Tutor
Joined: 27 Oct 2017
Last visit: 14 Nov 2025
Posts: 1,924
Own Kudos:
6,646
 [1]
Given Kudos: 241
WE:General Management (Education)
Expert
Expert reply
Posts: 1,924
Kudos: 6,646
 [1]
If a and b are positive integers, is a^4-b^4 divisible by 4? 06 Oct 2018, 10:25
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post

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
User avatar
pk123
User avatar
Joined: 16 Sep 2011
Last visit: 26 Apr 2020
Posts: 109
Own Kudos:
119
Given Kudos: 158
Products:
My Reviews
Posts: 109
Kudos: 119
If a and b are positive integers, is a^4-b^4 divisible by 4? 06 Oct 2018, 10:29
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
avatar
Manvi_nagpal
avatar
Joined: 09 Oct 2017
Last visit: 12 Feb 2019
Posts: 10
Own Kudos:
5
Given Kudos: 77
Location: India
Posts: 10
Kudos: 5
If a and b are positive integers, is a^4-b^4 divisible by 4? 06 Oct 2018, 10:31
Kudos
Add Kudos
Bookmarks
Bookmark this Post
A
Attachments

image.jpg
image.jpg [ 1.63 MiB | Viewed 9656 times ]

avatar
hargun3045
avatar
Joined: 06 Feb 2018
Last visit: 29 Nov 2022
Posts: 16
Own Kudos:
4
Given Kudos: 5
Posts: 16
Kudos: 4
If a and b are positive integers, is a^4-b^4 divisible by 4? 06 Oct 2018, 10:35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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

D
User avatar
Gladiator59
User avatar
Joined: 16 Sep 2016
Last visit: 18 Nov 2025
Posts: 838
Own Kudos:
2,613
Given Kudos: 260
Status:It always seems impossible until it's done.
GMAT 1: 740 Q50 V40
GMAT 2: 770 Q51 V42
Products:
My Reviews
GMAT 2: 770 Q51 V42
Posts: 838
Kudos: 2,613
If a and b are positive integers, is a^4-b^4 divisible by 4? 06 Oct 2018, 10:36
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.

Best,
G

Posted from my mobile device
User avatar
rever08
User avatar
Joined: 21 Jul 2017
Last visit: 13 Jan 2020
Posts: 151
Own Kudos:
115
Given Kudos: 143
Location: India
Concentration: Social Entrepreneurship, Leadership
Schools: ISB '20 (WL) Oxford"20 IIMA PGPX"20 (WD) Kelley '21 EDHEC Sept 2019 Intake
GMAT 1: 660 Q47 V34
GPA: 4
WE:Project Management (Education)
Products:
My Reviews
Schools: ISB '20 (WL) Oxford"20 IIMA PGPX"20 (WD) Kelley '21 EDHEC Sept 2019 Intake
GMAT 1: 660 Q47 V34
Posts: 151
Kudos: 115
If a and b are positive integers, is a^4-b^4 divisible by 4? 06 Oct 2018, 10:53
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Answer D
Attachments

Screen Shot 2018-10-06 at 11.21.26 PM.png
Screen Shot 2018-10-06 at 11.21.26 PM.png [ 268.85 KiB | Viewed 9615 times ]

User avatar
u1983
User avatar
Current Student
Joined: 24 Aug 2016
Last visit: 06 Jun 2021
Posts: 710
Own Kudos:
851
Given Kudos: 97
GMAT 1: 540 Q49 V16
GMAT 2: 680 Q49 V33
Products:
My Reviews
GMAT 2: 680 Q49 V33
Posts: 710
Kudos: 851
If a and b are positive integers, is a^4-b^4 divisible by 4? 06 Oct 2018, 10:59
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
OJA
User avatar
Joined: 21 Aug 2024
Last visit: 18 Nov 2025
Posts: 35
Own Kudos:
6
Given Kudos: 298
Location: India
Schools: HEC Sept '26
GPA: 4
Products:
GMAT Club Tests
Schools: HEC Sept '26
Posts: 35
Kudos: 6
If a and b are positive integers, is a^4-b^4 divisible by 4? 19 Feb 2025, 11:48
Kudos
Add Kudos
Bookmarks
Bookmark this Post
This is a nice question.
User avatar
anirchat
User avatar
Joined: 30 Jun 2024
Last visit: 14 Nov 2025
Posts: 288
Own Kudos:
44
Given Kudos: 323
Posts: 288
Kudos: 44
If a and b are positive integers, is a^4-b^4 divisible by 4? 19 Feb 2025, 11:49
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Agreed ! The intuition is simply great.
OJA
This is a nice question.
Show more
Moderators:
Bunuel
Math Expert
105355 posts
kingbucky
496 posts