Find all the ways of writing N into x^2-y^2

[ver_mathstats]

Homework Statement

When N=112, find the ways in which we can write it as x^2-y^2.

Relevant Equations

x^2-y^2, a^2 = b^2 mod(N)

We are required to find every way in which we can write 112 as x²-y². I already found one way using a
²≡b²modN. The values 11²≡3²mod(112) work and we can factor it as 14x8=112, I am confused how to approach this further for determining other values or would I just stop here? If I do test more values how exactly would I know when to stop? Any help would be appreciated, thank you!

 

[fresh_42]

Homework Statement:: When N=112, find the ways in which we can write it as x^2-y^2.
Relevant Equations:: x^2-y^2, a^2 = b^2 mod(N)

We are required to find every way in which we can write 112 as x²-y². I already found one way using a
²≡b²modN. The values 11²≡3²mod(112) work and we can factor it as 14x8=112, I am confused how to approach this further for determining other values or would I just stop here? If I do test more values how exactly would I know when to stop? Any help would be appreciated, thank you!

Whenever you see $x^2-y^2$ you can automatically write $(x-y)(x+y).$ Here we have the task to write
$$
N = (x+y)(x-y) = 2\cdot 2 \cdot 2 \cdot 2 \cdot 7
$$
Hence, we have only two prime divisors $p$ of a product. What is a prime divisor? A number $p$ is called prime if $p\,|\,a\cdot b$ implies $p\,|\,a$ or $p\,|\,b.$ This is all you need.

 

[SammyS]

Homework Statement:: When N=112, find the ways in which we can write it as x^2-y^2.
Relevant Equations:: x^2-y^2, a^2 = b^2 mod(N)

We are required to find every way in which we can write 112 as x²-y². I already found one way using a
²≡b²modN. The values 11²≡3²mod(112) work and we can factor it as 14x8=112, I am confused how to approach this further for determining other values or would I just stop here? If I do test more values how exactly would I know when to stop? Any help would be appreciated, thank you!

A complete and coherent statement of the problem would be much appreciated.

 

[ver_mathstats]

A complete and coherent statement of the problem would be much appreciated.

it is written completely and coherently :)

 

[Mark44]

it is written completely and coherently :)

Hmmm...

Homework Statement:: When N=112, find the ways in which we can write it as x^2-y^2.
Relevant Equations:: x^2-y^2, a^2 = b^2 mod(N)

We are required to find every way in which we can write 112 as x²-y²
. I already found one way using a 2≡b² modN. The values 11²
≡3² mod(112) work and we can factor it as 14x8=112

As far as the problem statement is concerned, I believe a clearer statement would be "Find all pairs of values x and y for which $x^2 - y^2 = 112$." As written, the problem statement doesn't say anything about equivalence classes, so I don't see how modular arithmetic plays a role in this problem.
Regarding your solution, $11^2 - 3^2 = 112$, so x = 11 and y = 3. x + y = 14 and x - y = 8, so (x + y)(x - y) = 112, as required.

I see at least one more pair of factors of 112 that satisfy $(x + y)(x - y) = 112$.

 

[PeroK]

Hmmm...As far as the problem statement is concerned, I believe a clearer statement would be "Find all pairs of values x and y for which $x^2 - y^2 = 112$."

I assumed it must be asking for pairs of (positive) integers

 

[ver_mathstats]

Hmmm...As far as the problem statement is concerned, I believe a clearer statement would be "Find all pairs of values x and y for which $x^2 - y^2 = 112$." As written, the problem statement doesn't say anything about equivalence classes, so I don't see how modular arithmetic plays a role in this problem.
Regarding your solution, $11^2 - 3^2 = 112$, so x = 11 and y = 3. x + y = 14 and x - y = 8, so (x + y)(x - y) = 112, as required.

I see at least one more pair of factors of 112 that satisfy $(x + y)(x - y) = 112$.

Yes, I found the other solution.

 

[WWGD]

Well, you can always look for the pair with :
x+y=N
x-y=1.

 

[SammyS]

Yes, I found the other solution.

Using the prime factorization (Thanks @fresh_42 .) and the hints by @Mark44 , I found two pairs of integer solutions, in addition to the pair: $(11,\ 3)$ .