Around the World in 80 Questions (Day 7): If a and b are integers and

If a and b are integers and a^2 - b^2 = 64, how many ordered pairs of (a, b) are possible ?
(a+b)(a-b)=64
a+b = 8, a-b = 8
then (a,b) = (6,2) 1st pair
a+b=-8, a-b=-8 then a = -8, b=0 2nd pair
a+b=16, a-b=4 then a=10,b=6 3rd pair
a+b=-16, a-b=-4 then a=-10,b=-6 4th pair
a+b=32, a-b=2 then a=17, b=15 5th pair
a+b=-32, a-b=-2 then a =-34, b= -2 6th pair

answer is 6 pairs

My Answer: B

Ordered pairs satisfying this equation are: (8,0), (-8,0), (10,6),(-10,6), (-10,-6), (10,-6), (17,15), (17,-15), (-17,15), (-17,-15).

Hence, 10 ordered pairs are there.

\(a^{2}-b^{2}= (a+b) (a-b)= 64 \), so firstly we will just break and see factors so number of factors for this is 7
the\( ( a+b ) (a-b ) \)can have following values
64*1
2*32
4*16
8*8
16*4
32*2
1*64

one thing to note here is if a+b and a-b add upto an odd number which happens when a+b=64 and a-b=1 then value of a is always a fraction in this case 65/2 so we rule out those 2 options 64*1 and 1*64

All other options give us unique values of a and b like in case of 2* 32 or 32 * 2 we get values as (17, 15 ) and (17,-15 ) respectively so the answer I marked is 5.

The answer is D= 5
The factor pairs for 64 are- (1,64), (2, 32), (4,16), (8,8),
(1, 64) does not give any integer solution
(2, 32) gives (a+b)=32 and (a-b)=2, by adding the 2 solutions we get 2a=34 , a=17 and b=15,
Hence (17, 15 ) is a valid solution.
For the factor pair (4,16) (a + b) = 16 and (a - b) = 4
Adding the two equations: 2a = 20
a = 10, so b = 6
The pair (a, b) = (10, 6) is a valid solution.
For factor pair (8,8) (a + b) = 8 and (a - b) = 8
Adding the two equations: 2a = 16
a = 8, so b = 0
The pair (a, b) = (8, 0) is a valid solution.
(a + b) = 4 and (a - b) = 16
Adding the two equations: 2a = 20
a = 10
Substituting back: 10 + b = 4, so b = -6
The pair (a, b) = (10, -6) is a valid solution.
For factor pair (2,32) (a + b) = 2 and (a - b) = 32
Adding the two equations: 2a = 34
a = 17, so b = -15
The pair (a, b) = (17, -15) is a valid solution.
So, the valid ordered pairs (a, b) that satisfy the equation a^2 - b^2 = 64 are:

(17, 15), (10, 6), (8, 0), (10, -6), and (17, -15) Hence 5 pairs possible

8,0
-8,0
10,6
-10,6
-10,-6
10,-6
total are 6

(a+b)(a-b) = 64
Since a,b are integers, a+b and a-b are integers as well.
So, we just need to find factors of 64 - 1,2,4,8,8,16,32,64
Taking 2*32 combination,
a+b = 32
a-b = 2
gives us a = 17, b = 15
Since it is a^2 - b^2 = 64, a= -17, b= -15 also satisfy the equation.
a can take +17, -17 and b can take +15, -15. So 4 pairs are possible.

Same is the case for 4*16 combination. So far, we have 8 solutions.

In case of 8*8,
a+b = 8
a-b = 8
The solution is a = +8 or -8 and b = 0. So, we can get only 2 solutions.
So far, we have 10 solutions.

In case of 1*64,
a+b = 64
a-b = 1
a = 65/2 which is not integer. So, this case yields no solution.

So, we have 10 solutions. B is the answer.

Asked: If a and b are integers and a^2 - b^2 = 64, how many ordered pairs of (a, b) are possible ?

a^2 - b^2 = 64
(a+b)(a-b) = 64

Case 1:
a + b = 32; a - b = 2
a = 17; b = 15; (a, b) = (17,15)
Other 3 cases are also possible
a = -17; b = 15; (a, b) = (-17,15)
a = 17; b = -15; (a, b) = (17,-15)
a = - 17; b = - 15; (a, b) = (-17,-15)
4 ordered pairs

Case 2:
a + b = 16; a - b =4
a = 10; b = 6; (a, b) = (10,6)
Other 3 cases are also possible
a = -10; b = 6; (a, b) = (-10,6)
a = 10; b = -6; (a, b) = (10,-6)
a = -10; b= - 6;; (a, b) = (-10,-6)
4 ordered pairs

Case 3:
a + b = 8; a - b = 8
a = 8; b = 0; (a, b) = (8,0)
Other 1 case is also possible
a = -8; b = 0; (a, b) = (-8,0)
2 ordered pairs

Total ordered pairs = 4+ 4+ 2 = 10 ordered pairs

IMO B

a^2 - b^2 = 64

(8,0), (-8,0), (10,6), (-10,6), (10,-6), (-10,-6), (17,15), (-17,15), (17,-15), (-17,-15)

Ans = 10

Ans B

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(a-b)(a+b)=64.1,-64.-1, 31.2, -32.-2, 2.32,-2.-32...8.8,-8,-8
ignore 64.1 (odd sol)

(a-b)(a+b)= (±32,±2), (±2,±32),(±16,±4),(±4,±16), (±8,±8)
a=sum/2, b=diff/2
a=±17 ,±10, ±8 , b=±15 , ±6, 0

10 ordered solutions

B)

possible pairs of a,b for
a^2 - b^2 = 64

(8,0) ; ( -8,0) ; ( 10,6) ; ( -10,-6) , ( 10,-6) ; (-10,6) ; ( 17,15); ( -17,-15); (17,-15) ;(-17,15)
total such pairs are 10
option B

6 ordered pairs can satisfy a^2-b^2=64

(8,0) and (-8,0), which would result in +/- 8^2- 0^2 =64
(10,6), (-10,6), (10,-6) and (-10,-6) can also satisfy the equation to equal 64

(a+b)(a-b) = 64

1) Both (a+b) and (a-b) are positive.
2) Both (a+b) and (a-b) are negative

The possible value of (a,b) are shown below. Total 10 pairs.



We ignore two pairs as the value of a will not be an integer.

IMO B

(a+b)(a-b) = 64
64 = (1*64); (2*32); (4*16); (8*8); (-1*-64); (-2*-32); (-4*-16); (-8*-8)

1) 64 = 1*64
If a-b = 1 => a= 1+b
=> Is a+b = 64? => 1+b+b = 64? => 1+2b= 64? => 2b = 63? => b is not an integer so not possible

2) 64 = 2*32
If a-b = 2 => a= 2+b
=> Is a+b = 32? => 2+b+b = 32? => 2+2b= 32? => 2b = 30? => b = 15 ---> YES
Possible sets = 2*32; 32*2; -2*-32; -32*-2 [#4 sets]

3) 64 = 4*16
If a-b = 4 => a= 4+b
=> Is a+b = 16? => 4+b+b = 16? => 4+2b= 16? => 2b = 12? => b = 6 ---> YES
Possible sets = 4*16; 16*4; -4*-16; -16*-4 [#4 sets]

4) 64 = 8*8
If a-b = 8 => a= 8+b
=> Is a+b = 8? => 8+b+b = 8? => 2b= 0? => b = 0 ---> YES
Possible sets = 8*0; 0*8; -8*0; 0*-8 [#4 sets]

Total pairs = 4+4+4 = 12 [A]

Solution is given below

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Only thing to keep in mind is that a and b can be negative as well.

a^2 - b^2 = 64
(a-b)(a+b) = 2^6

(a+b) can be 64, 32, 16, 8, 4, 2, 1, -1, -2, -4, -8, -16, -32, -64. Correspondingly,
(a-b) can be 1, 2, 4, 8, 16, 32, 64, -64, -32, -16, -8, -4, -2, -1.

Out of these solutions (a,b) = (17,15), (10,6) and (8,0).
But a, b can be negative too, so (+-17,+-15), (+-10,+-6) and (+-8,0).

Total 10 ordered pairs possible.

IMO B.

If a and b are integers and a^2 - b^2 = 64, how many ordered pairs of (a, b) are possible ?

A. 12
B. 10
C. 6
D. 5
E. 3
___________________
Basically, this question can be solved by placing numbers instead of A and B , but it could be time consuming.
For example. A =10 and B=6 gives us 64. Still, we need a more efficient way to solve it . Expert's reply.

a^2-b^2=64
(a-b)(a+b)=64
since a and b are integers, they could have following respective values: (2,32), (4,16), (8,8), (32,2), (16,4)
Hence 5. D is correct.