Algebra Problem

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.

#Algebra #HelpMe! #MathProblem

Easy Math Editor

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Comments

Note 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.

Sreejato Bhattacharya - 8 years ago

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 $0 \leq x \leq 9$ ).

Calvin Lin Staff - 8 years ago

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

Bob Krueger - 8 years ago

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).

Nishant Sharma - 8 years ago

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

Sreejato Bhattacharya - 8 years ago

how much u getting in iit

A Former Brilliant Member - 8 years ago

@A Former Brilliant Member – Are you talking about ADVANCED ? Not good at all. It's 150 only ? How about you ? And what's your score in MAINS ?

Nishant Sharma - 8 years ago

@Nishant Sharma – u'll score b/w 5000-7000 i guess....

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

A Former Brilliant Member - 7 years, 12 months ago

@A Former Brilliant Member – Hey don't be so depressed. You must be getting somewhere else a good rank( I mean you must have appeared in exams like BITSAT, VITEEE and so on). Always keep hope. Even I don't think I would be getting into a good institute with this score.

Nishant Sharma - 7 years, 12 months ago

@Nishant Sharma – but atleast you'll get ism,or it bhu or rgipt but i won't even get that..... :'( ya may be i'll get them....but iit was the dream nd it shattered.....

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

A Former Brilliant Member - 7 years, 12 months ago

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

Triptesh Biswas - 7 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$