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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.
y^2 = x^3 + n, n is some integer.
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.robert Ihnot said: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.
robert Ihnot said: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: [tex]1= (b^2+ab+a^2)\frac{b-a}{4}[/tex] Clearly b-a can not be zero and with only positive terms a^2+ab+b^2 exceeds 1.
al-mahed said: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 said:why did you conclude that x-2 and x+2 can each be written as a cube?
CRGreathouse said: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 said:There are no cubes which are 4 apart. The difference between consecutive cubes grows rapidly (quadratically).
An integer solution is a set of values for x and y that satisfy the given equation and are both whole numbers (positive, negative, or zero).
To find integer solutions for this equation, you can use a method called "trial and error." Start by plugging in different values for x and solving for y. If the resulting y value is a whole number, then you have found an integer solution. If not, try a different value for x until you find a solution.
Yes, there can be multiple integer solutions for this equation. In fact, there are often infinitely many solutions, as there are infinite values for x and y that satisfy the equation.
Yes, there are some patterns and shortcuts that can be used to find integer solutions for this equation. For example, if n is a perfect square, then there will be integer solutions for x and y. Also, if n is a multiple of a perfect square, there may be more than one integer solution. Additionally, some integer solutions can be found by factoring the equation or using modular arithmetic.
Yes, there are computer programs and online tools available that can help find integer solutions for this equation. However, it is still important to understand the underlying concepts and methods used to find these solutions.