68068 as difference of two squares

[Robert Houdart]

Homework Statement

In how many ways can 68068 be written as the difference of two squares?

Homework Equations

The Attempt at a Solution

Let (x+a) * (x+a) -x*x =68068=2*2*7*11*13*17
a (2x+a) =2*2*7*11*13*17
As 2x+a is odd ⇒ a is even
∴a=2b
2b (2x+2b) =2*2*7*11*13*17
b (x+b) =7*11*13*17
x= (7*11*13*17) /b - b
Since 7*11*13*17 has 16 factors ∴ x has 16 different values
However, the book states 8 as the answer.

 

[RUber]

It looks like you need to account for the pairs (x,b) (b,x) , since those would not be distinct differences of squares.

 

[Robert Houdart]

i

It looks like you need to account for the pairs (x,b) (b,x) , since those would not be distinct differences of squares.

I guess you are right. (a, b) and (b, a) are repeated in my solution, taking pair (a,b) and (b,a) as a single pair yields 8 as the answer.

 

[PeroK]

i

I guess you are right. (a, b) and (b, a) are repeated in my solution, taking pair (a,b) and (b,a) as a single pair yields 8 as the answer.

That's not quite the reason. Remember that you had factors b and x+b. That implies a constraint.

 

[Rochette_rocket]

Could you please explain why 2x+a is odd

 

[Mark44]

Could you please explain why 2x+a is odd

If you're talking about these lines in post #1:

a (2x+a) =2*2*7*11*13*17
As 2x+a is odd ⇒ a is even

We have $a(2x + a)$ being equal to an even number.
If a were odd, then 2x + a would also be odd, and we would have an odd integer times another odd integer, which can't possibly multiply to make an even integer.
Therefore, a must be even.
All of this can be proven more rigorously than how I've stated things.