What are all positive integral solutions (x,y) of y³ = x³ + 11x² - 5x + 27?

Question

The proof that it is all of them is the important part.

Answer 1

Since (x+4)^3 > x^3 + 11x^2 -5x +27 = f(x) , we test

f(x) = (x+3)^3 and find equality at x=16 and x=0 giving solutions

(x,y) = (16, 19) and (0,3)

f(x) = (x+2)^3 gives 5x^2 = 17x - 19 forcing x = 19

a contradiction. Also f(x) = (x+1)^3 fails mod 8:

8x^2 = 8x - 26. Leaving f(x) = x^3 or 11x^2 -5x +27 = 0

but 11x^2-5x +27 >0 for x>=2 i.e., f(x) > x^3 for x>=2

and fails for x=1. Finally f(x)> (x-1)^3 for all natural x.

So, the only solutions are (x,y) = (0,3) and (16,19)

Answer 2

A partial solution: will update more if I think of something.

If (x,y) is an integral solution, we can set x=z+1 where z is an integer. Plug it in we get (z,y) is an integer solution of

y³ = (z+1)³ + 11(z+1)² - 5(z+1) + 27 = z³ + 14z² + 20z + 34

Suppose now that y is a even number, so y³ is divisible by 8: write it as 8w. So we have

z³ + 14z² + 20z + 2(17-4w) = 0

Notice that 17-4w is an odd number. Now we can apply Eisenstein's criterion by seeing that 2 divides the last three coefficients but 4 does not divide the constant term, and so the polynomial is irreducible (not factorable) and cannot have an integer solution. This proves that there cannot be integral solutions to the equation where y is even.

Answer 3

I tryed numerous the thank you to attempt fixing this equation and the only answer i ought to arise with is: {(3x+2y,5x+y,x,y)|x,yER} The E is that curve with the line interior the midsection and the R is amazingly formidable. i don't understand who to make those symbols on the computing device. i don't think of it incredibly is the respond you're searching for in spite of the fact that it exchange into all i ought to arise with. i in my view think of this equation won't be solvable.