Finding Integer Solutions for y^2 = x^3 + n

[quila]

how does one go about finding integer solutions for an equation such as this? Is it easier to merely find how many solutions?

y^2 = x^3 + n, n is some integer.

 

[al-mahed]

if n = +- 2 this is a classic Fermat's problem: prove that 26 is the only natural number between an square and a cube (5^2 = 25 and 3^3 = 27)

 

[quila]

what if n is large? say n>100

 

[al-mahed]

where did you find this problem? Could you put here with the exacly words?

 

[robert Ihnot]

I thought X^2=Y^3 + 1 was a well-known problem.

There are trivial solutions, such as y=0, x=1, or x=0, y=-1. Ingnoring that, we have (X-1)(X+1) = y^3, with all factors assumed positive. X-1 and X+1 can only have 2 as a common factor. And, in that case, one term is divisible by 4. X-1=2a^3, X+1 = 4b^3, assuming a>0, b>0, this gives us 2=4a^3-2b^3, or 1 =2a^3-b^3. This is only possible if a=b=1. Thus the solution is 3^2=2^3+1.

For negative terms,b=-1, a=-1, 1=b^3-2a^3, giving x-1=4a^3=-4, x+1=2b^3=-2, we could have found, but excluded: (-3)^2= 2^3+1 for a=b=-1.

If they have nothing in common we get 2 = b^3-a^3, which gives no answer in positive integers. However, X-1= a^3=-1, X+1= b^3=1, gives X=0, Y=-1, another trivial result.

Thus there is only one non-trivial solution, already given: 3^2= 2^3+1.

 

[ramsey2879]

I thought X^2=Y^3 + 1 was a well-known problem.

There are trivial solutions, such as y=0, x=1, or x=0, y=-1. Ingnoring that, we have (X-1)(X+1) = y^3, with all factors assumed positive. X-1 and X+1 can only have 2 as a common factor. And, in that case, one term is divisible by 4. X-1=2a^3, X+1 = 4b^3, assuming a>0, b>0, this gives us 2=4a^3-2b^3, or 1 =2a^3-b^3. This is only possible if a=b=1. Thus the solution is 3^2=2^3+1.

For negative terms,b=-1, a=-1, 1=b^3-2a^3, giving x-1=4a^3=-4, x+1=2b^3=-2, we could have found, but excluded: (-3)^2= 2^3+1 for a=b=-1.

If they have nothing in common we get 2 = b^3-a^3, which gives no answer in positive integers. However, X-1= a^3=-1, X+1= b^3=1, gives X=0, Y=-1, another trivial result.

Thus there is only one non-trivial solution, already given: 3^2= 2^3+1.

That is the solution where n = 1, but more importantly what the poster asks is what method can be used to obtain the solution if n = other, e.g. 22. One could pose an infinite number of such questions but I doubt that there is a direct methof of solution. At least there is no discussion of such a problem as far as I know.

 

[al-mahed]

there are infinite n such that n = x^2 - y^3, this is trivial

there are at least one x and one y such that n = x^2 - y^3 for every n integer? this is the question, I think... it's hard!

 

[robert Ihnot]

ramsey2879: At least there is no discussion of such a problem as far as I know.

I agree. I don't think anything has been done beyond piecemeal efforts. Obviously, n=1 is a simple case.

al-mahed: if n = +- 2 this is a classic Fermat's problem: prove that 26 is the only natural number between an square and a cube (5^2 = 25 and 3^3 = 27)

I have some memory of 3^3=5^2+2. I can't seem to find any information about these problems at all.

 

[robert Ihnot]

al-mahed: there are at least one x and one y such that n = x^2 - y^3 for every n integer? this is the question, I think... it's hard!
That, I feel is not true. Take 4=X^2-Y^3. Obviously Y=0 is such a solution, a negative Y impossible. But for X and Y greater than 0, consider them both even Then in this case we have 4=4x'^2+8y'^2. This leaves us 1=x'^2+2y'^3. Only solution y' = 0.

Now if X and Y are both odd,Y^3=(X-2)(X+2), assume X>2, these have no common factor, (since only 2 could be such a factor) x-2 =a^3, x+2 = b^3. This then gives
4 = b^3-a^3 = (b-a)(b^2+ab+a^2). Since a and b are odd, 4 divides b-a, and thus we have the form: $$1= (b^2+ab+a^2)\frac{b-a}{4}$$ Clearly b-a can not be zero and with only positive terms a^2+ab+b^2 exceeds 1.

 

[Xevarion]

Just so you know, this is a hard problem in general...
http://en.wikipedia.org/wiki/Elliptic_curve#Connections_to_number_theory

 

[al-mahed]

al-mahed: there are at least one x and one y such that n = x^2 - y^3 for every n integer? this is the question, I think... it's hard!
That, I feel is not true. Take 4=X^2-Y^3. Obviously Y=0 is such a solution, a negative Y impossible. But for X and Y greater than 0, consider them both even Then in this case we have 4=4x'^2+8y'^2. This leaves us 1=x'^2+2y'^3. Only solution y' = 0.

Now if X and Y are both odd,Y^3=(X-2)(X+2), assume X>2, these have no common factor, (since only 2 could be such a factor) x-2 =a^3, x+2 = b^3. This then gives
4 = b^3-a^3 = (b-a)(b^2+ab+a^2). Since a and b are odd, 4 divides b-a, and thus we have the form: $$1= (b^2+ab+a^2)\frac{b-a}{4}$$ Clearly b-a can not be zero and with only positive terms a^2+ab+b^2 exceeds 1.

I did not follow you here: Now if X and Y are both odd,Y^3=(X-2)(X+2), assume X>2, these have no common factor, (since only 2 could be such a factor) x-2 =a^3, x+2 = b^3.

why did you conclude that x-2 and x+2 can each be written as a cube?

 

[CRGreathouse]

why did you conclude that x-2 and x+2 can each be written as a cube?

He took n = 4, which means 4 = X^2-Y^3. Adding Y^3 - 4 to each side gives Y^3 = X^2 - 4 = (X - 2)(X + 2). Since they have no common factors, if one is not a cube then the product is not a cube.

 

[Xevarion]

why did you conclude that x-2 and x+2 can each be written as a cube?

There are no cubes which are 4 apart. The difference between consecutive cubes grows rapidly (quadratically).

 

[al-mahed]

He took n = 4, which means 4 = X^2-Y^3. Adding Y^3 - 4 to each side gives Y^3 = X^2 - 4 = (X - 2)(X + 2). Since they have no common factors, if one is not a cube then the product is not a cube.

Now I see, thanks!

I made my question because I had conclude that if only one is a cube the product can not be a cube, but is you right when you say: both need to be a cube, and if no one is a cube, there is no way an odd number N with the same factor than N + 4

 

[al-mahed]

There are no cubes which are 4 apart. The difference between consecutive cubes grows rapidly (quadratically).

Yes, this part is clear, I made another question, concerning the fact that y^3=(x-2)(x+2) ==> y| (x-2) or (x+2) only if y is prime (of course y dividing only one makes impossible the product be a cube). If y is not prime was not so clearly for me that (x-2)(x+2) cannot form a cube, although to be almost obvious.

 

[robert Ihnot]

What ever prime divides (x-2) and (x+2) also divides the sum, which is 4.

The fundamental theorem of arithmetic tells us that any integer can be decomposed into prime factors in only one way, without regard for order. Thus we only need to consider prime factors.