Solve the equation \(y^{3}\) = \(x^{3}\) + 8\(x^{2}\) - 6\(x\) + 8 for positive integers x and y.

I solved it this way:

\(y^{3}\) - 8 = \(x^{3}\) + 8\(x^{2}\) - 6\(x\)

(\(y\) - 2)(\(y^{2}\) + 4 + 2y) = \(x\)(\(x^{2}\) + 8\(x\) - 6)

Then one possible solution came from equating the like positioned factors on each side. Since second factor on L.H.S cannot be factorized further and the last factor on R.H.S after further factorization involves irrational constants, so another only possible solution would come from equating 1st(L.H.S) and 2nd(R.H.S) factors, 2nd(L.H.S) and 1st(R.H.S) which on simplification gives an equation involving all positive terms which will obviously yield no positive solutions.

I would appreciate if someone gives me a method better than this to solve this problem.

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## Comments

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TopNewestNote that if a quadratic polynomial of the form \( ax^2 + bx + c \) (\( a > 0\))has a negative discriminant then it will always be positive for all real values of \( x \). Now we first prove that \( y > x \). This is true because the polynomial \( 8x^2 - 6x + 8 \) has a negative discriminant ( \( -220 \) ), so it is always positive. Thus \( y^3 > x^3 \implies y> x \). Now we examine if \( y= x+1 \) is possible. Consider \( y^3 - (x+1)^3 = 8x^2 - 6x + 8 - (x+1)^3= 5x^2 - 9x + 7 \). This polynomial has a negative discriminant \( - 59 \), which means that it is positive for all values of \( x \), thus \( y^3 > (x+1)^3 \implies y>x+1\). Now we will see if \( y = x+2 \) is possible. Note that \( y^3 - (x+2)^3 = 2x^2 - 18x= 2 [(x-\frac{9}{2})^2 - \frac{81}{4}] \). Thus this has a minimum value of \( -\frac{81}{4} \) and no maximum value. So \( y^3 - (x+2)^3 \) can be \( 0 \). Now we examine if \( y= x+3 \) is possible. Note that \( (x+3)^3 - y^3= x^2 + 33x + 19 \). Note that the function \( x^2 + 33x + 19 \) is monotonically increasing over the positive reals, and since \( x \) is a positive integer, the minimum possible value for \( x \) is \( 1 \), which implies minimum possible value of \( x^2 + 33x + 19 \) is \( 1+33+19 = 53 > 0 \). Thus we have \( (x+3)^3 > y^3 \implies x+3>y \). So combining the previously proved results we have \( x+1 < y < x+ 3 \). Since \( y \) is an integer the only possible value for \( y \) is \( x+ 2 \). Thus we have \( y^3 - (x+2)^3= 0 \) which reduces to \( 2x^2 - 18x = 0 \implies x= 9 \) [since \(x>0\)] For \( x= 9 \) we have \( y= 11\). Hence \( (x, y)= (9, 11) \) is the only possible solution over the positive integers.

I think you made a mistake in assuming that the numbers \( x^2 + 8x - 6 \) and \( y^2 +2y + 4 \) are always prime since they are algebraically irreducible. However [though it does not satisfy the equation], if you take \( x= 2 \) [or any even number] you will see that \( x^2 + 8x - 6 \) is a composite number, and similarly if you take \( y= 2 \) [or any even number] you will see that \( y^2 + 2y + 4 \) is a composite number. If a polynomial \( p(x) \) can be factorized over integer coefficients then \( p(x) \) will always be a composite number for all integer values of \( x \) (except when the factors of \( p(x) \) are equal or one of the factors is \( 0 \) for some value of \( x \)), however if \( p(x) \) cannot be factorized over integer coefficients it doesn't mean that \( p(x) \) will always be prime for all integer values of \( x \). This is what I think. Please correct me if I am wrong.

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In a similar train of thought, for all but finitely many integers \(x\), we have \( (x+1)^3 < y^3 < (x+2)^3 \). Hence we just need to check those cases (which give \( 0 \leq x \leq 9 \)).

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Yeah, I understood my mistake. I could not have thought of doing things like you have done to ultimately bound y. I just took it as a factorization problem and proceeded thereupon. Now I get your idea which is even though a bit lengthy but surely error-free. That was nicely explained, except for the solution which I think would be (9,11).

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Yes the value of \( y \) is \( 11 \), not \( 11^3 \). I forgot to take the cube root. Very sorry for the silly mistake.

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how much u getting in iit

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forget it m not gonna clear advance i think..... :(

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can u help me... over a matter????????? plz

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Flawless explaination except for one part: since x=9, y=11. You had y equalling \(11^3\).

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X=9 and Y=11 is the solution of this problem

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