how to solve this equation with only positive integer solutions, X^2 + Y^3 = 3?

Question

even if you could just let me know how to get started it would be appreciated

Answer 1

To solve it with only positive integer solutions, try positive integers less than 3. There are only 2 of them, 1 and 2, so try various combinations and you'll see that there are no positive integer solutions to this. However, if you allow negative integer solutions, you can also try -1 and -2 for x and y. Then you'll find that x can be + or - 2, and y would be -1.

There are infintiely many solutions to this equation but most of them are irrational numbers, such as x = 0 and y = cube root of 3, y = 0 and x = sqrt 3, etc. Try graphing this equation on your graphics calculator and see what it looks like.

Answer 2

constructive integers are also commonplace because the organic numbers: a million, 2, 3, ... What are the powers of two for organic exponents? 2^a million = 2 2^2 = 4 2^3 = 8 2^4 = 16 2^5 = 32 and so on. What are the powers of three for organic exponents? 3^a million = 3 3^2 = 9 3^3 = 27 3^4 = 80 one 3^5 = 243 3^6 = 729 enable's go by using the list of three^y and upload 5 to each. 3^a million + 5 = 8 = 2^3 3^2 + 5 = 14 3^3 + 5 = 32 = 2^5 3^4 + 5 = 86 3^5 + 5 = 248 3^6 + 5 = 734 properly, (3, a million) and (5, 3) are recommendations, yet can we stumble on any more desirable or educate that there aren't any more desirable?

Answer 3

If x=1, y^3=2 and y=2^(1/3)

If y=1, x^2=2 and x=2^(1/2)

Answer 4

IF you look at nightshade's answer, you have x in terms of y

Just start by isolating the variables:
x^2+y^3(-y^3)=3-y^3
x^2=3-y^3
x=Sqrt(3-y^3)

So, for integer solutions, you need a perfect square of the form 3-y^3. the y values will need to be negative (domain of sqrt is positive).

Starting with y=-1, you get a nice x value of 2.

Now you just need to prove this is the only solution!

Answer 5

There are no POSITIVE integer solutions:
Rewrite this as x² = 3-y^3.
Since y is positive, this says x² < 3,
so x = 1. But x = 1 doesn't furnish a solution!
A much greater challenge is to find all INTEGER
solutions(positive or negative).

Answer 6

there is more then 1 why to get positive number but it will be long. 1st is set x to zero or set y to zero that will be the first step but if you want both x & y to have numbers just do x^5 hope that could give you a start

Answer 7

Just start by isolating the variables:
x^2+y^3(-y^3)=3-y^3
x^2=3-y^3
x=Sqrt(3-y^3)

x^2(-x^2)+y^3=3-x^2
y^3=3-x^2
y=Cube Rt(3-x^2)

Answer 8

x=2
y=-1
2^2+(-1)^3=4-1=3