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

\((a^4 - b^4\)) % 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)

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?

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.

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

Can anyone please explain why the remainder = 2, a^2 + b^2 both are odd or both are even?[/quote]

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

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

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

A

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

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

Answer D

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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