Algebra Problem

Solve the equation y3y^{3} = x3x^{3} + 8x2x^{2} - 6xx + 8 for positive integers x and y.

I solved it this way:

y3y^{3} - 8 = x3x^{3} + 8x2x^{2} - 6xx

(yy - 2)(y2y^{2} + 4 + 2y) = xx(x2x^{2} + 8xx - 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.

Note by Nishant Sharma
6 years, 4 months ago

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5 votes

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Note that if a quadratic polynomial of the form ax2+bx+c ax^2 + bx + c (a>0 a > 0)has a negative discriminant then it will always be positive for all real values of x x . Now we first prove that y>x y > x . This is true because the polynomial 8x26x+8 8x^2 - 6x + 8 has a negative discriminant ( 220 -220 ), so it is always positive. Thus y3>x3    y>x y^3 > x^3 \implies y> x . Now we examine if y=x+1 y= x+1 is possible. Consider y3(x+1)3=8x26x+8(x+1)3=5x29x+7 y^3 - (x+1)^3 = 8x^2 - 6x + 8 - (x+1)^3= 5x^2 - 9x + 7 . This polynomial has a negative discriminant 59 - 59 , which means that it is positive for all values of x x , thus y3>(x+1)3    y>x+1 y^3 > (x+1)^3 \implies y>x+1. Now we will see if y=x+2 y = x+2 is possible. Note that y3(x+2)3=2x218x=2[(x92)2814] y^3 - (x+2)^3 = 2x^2 - 18x= 2 [(x-\frac{9}{2})^2 - \frac{81}{4}] . Thus this has a minimum value of 814 -\frac{81}{4} and no maximum value. So y3(x+2)3 y^3 - (x+2)^3 can be 0 0 . Now we examine if y=x+3 y= x+3 is possible. Note that (x+3)3y3=x2+33x+19 (x+3)^3 - y^3= x^2 + 33x + 19 . Note that the function x2+33x+19 x^2 + 33x + 19 is monotonically increasing over the positive reals, and since x x is a positive integer, the minimum possible value for x x is 1 1 , which implies minimum possible value of x2+33x+19 x^2 + 33x + 19 is 1+33+19=53>0 1+33+19 = 53 > 0 . Thus we have (x+3)3>y3    x+3>y (x+3)^3 > y^3 \implies x+3>y . So combining the previously proved results we have x+1<y<x+3 x+1 < y < x+ 3 . Since y y is an integer the only possible value for y y is x+2 x+ 2 . Thus we have y3(x+2)3=0 y^3 - (x+2)^3= 0 which reduces to 2x218x=0    x=9 2x^2 - 18x = 0 \implies x= 9 [since x>0x>0] For x=9 x= 9 we have y=11 y= 11. Hence (x,y)=(9,11) (x, y)= (9, 11) is the only possible solution over the positive integers.

I think you made a mistake in assuming that the numbers x2+8x6 x^2 + 8x - 6 and y2+2y+4 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 x= 2 [or any even number] you will see that x2+8x6 x^2 + 8x - 6 is a composite number, and similarly if you take y=2 y= 2 [or any even number] you will see that y2+2y+4 y^2 + 2y + 4 is a composite number. If a polynomial p(x) p(x) can be factorized over integer coefficients then p(x) p(x) will always be a composite number for all integer values of x x (except when the factors of p(x) p(x) are equal or one of the factors is 0 0 for some value of x x ), however if p(x) p(x) cannot be factorized over integer coefficients it doesn't mean that p(x) p(x) will always be prime for all integer values of x x . This is what I think. Please correct me if I am wrong.

Sreejato Bhattacharya - 6 years, 4 months ago

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In a similar train of thought, for all but finitely many integers xx, we have (x+1)3<y3<(x+2)3 (x+1)^3 < y^3 < (x+2)^3 . Hence we just need to check those cases (which give 0x9 0 \leq x \leq 9 ).

Calvin Lin Staff - 6 years, 4 months ago

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Flawless explaination except for one part: since x=9, y=11. You had y equalling 11311^3.

Bob Krueger - 6 years, 4 months ago

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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).

Nishant Sharma - 6 years, 4 months ago

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Yes the value of y y is 11 11 , not 113 11^3 . I forgot to take the cube root. Very sorry for the silly mistake.

Sreejato Bhattacharya - 6 years, 4 months ago

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how much u getting in iit

A Former Brilliant Member - 6 years, 4 months ago

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@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 - 6 years, 4 months ago

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

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

A Former Brilliant Member - 6 years, 4 months ago

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@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 - 6 years, 4 months ago

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@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 - 6 years, 4 months ago

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

Triptesh Biswas - 6 years ago

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