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

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These are: ##(57,\ 55)## and ##(15,\ 13)##.In summary, we are required to find all pairs of values x and y for which ##x^2 - y^2 = 112##. By using the prime factorization of 112, we can find other solutions besides the one given in the conversation. These include the pairs (57, 55) and (15, 13).
  • #1
ver_mathstats
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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!
 
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  • #2
ver_mathstats said:
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.
 
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  • #3
ver_mathstats said:
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.
 
  • #4
SammyS said:
A complete and coherent statement of the problem would be much appreciated.
it is written completely and coherently :)
 
  • #5
ver_mathstats said:
it is written completely and coherently :)
Hmmm...

ver_mathstats said:
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##.
 
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  • #6
Mark44 said:
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
 
  • #7
Mark44 said:
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.
 
  • #8
Well, you can always look for the pair with :
x+y=N
x-y=1.
 
  • #9
ver_mathstats said:
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)## .
 

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