Here is a challenging diophantine problem to ponder about.
2x3=x2y4+9y52x^{3}=x^{2}y^{4}+9y^{5}2x3=x2y4+9y5
6y3=3x3+xy36y^{3}=3x^{3}+xy^{3}6y3=3x3+xy3
Find x and y so that they are integer solutions*
Note by Kevin H 7 years, 7 months ago
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If both xxx and yyy are rational and nonzero, then r=xyr=\tfrac{x}{y}r=yx is rational and nonzero. Since 6y3=3x3+xy36y^3 = 3x^3 + xy^36y3=3x3+xy3, we deduce that x=6−3r3x = 6 - 3r^3x=6−3r3, and of course y=xr=6−3r3ry = \frac{x}{r} = \frac{6-3r^3}{r}y=rx=r6−3r3. From the first equation we deduce that 2x3=x2y4+9y52x3=x2x4r4+9x5r5 = x6r4+9x5r52=x3r4+9x2r52r5=rx3+9x2 = 27r(2−r3)3+81(2−r3)2 \begin{array}{rcl} 2x^3 & = & x^2y^4 + 9y^5 \\ 2x^3 & = & x^2 \frac{x^4}{r^4} + 9\frac{x^5}{r^5} \; = \; \frac{x^6}{r^4} + \frac{9x^5}{r^5} \\ 2 & = & \frac{x^3}{r^4} + \frac{9x^2}{r^5} \\ 2r^5 & = & rx^3 + 9x^2 \; = \; 27r(2 - r^3)^3 + 81(2-r^3)^2 \end{array} 2x32x322r5====x2y4+9y5x2r4x4+9r5x5=r4x6+r59x5r4x3+r59x2rx3+9x2=27r(2−r3)3+81(2−r3)2 and hence f(r) = 27r10−162r7−81r6+2r5+324r4+324r3−216r−324 = 0 f(r) \; =\; 27r^{10} - 162r^7 - 81r^6 + 2r^5 + 324r^4 + 324r^3 - 216r - 324 \; = \; 0 f(r)=27r10−162r7−81r6+2r5+324r4+324r3−216r−324=0 If we write r=abr = \frac{a}{b}r=ba where a,ba,ba,b are coprime integers with b≥1b \ge 1b≥1, then 27a10−162a7b3−81a6b4+2a5b5+324a4b6+324a3b7−216ab9−324b10 = 0 27a^{10} - 162a^7b^3 - 81a^6b^4 + 2a^5b^5 + 324a^4b^6 + 324a^3b^7 - 216ab^9 - 324b^{10} \; = \; 0 27a10−162a7b3−81a6b4+2a5b5+324a4b6+324a3b7−216ab9−324b10=0 Thus we deduce that b3b^3b3 divides 27a1027a^{10}27a10, and hence b3b^3b3 divides 272727, and hence b=1,3b = 1,3b=1,3. Moreover aaa divides 324b10324b^{10}324b10, and so aaa divides 324324324.
If b=3b=3b=3 then, since 324=4×81324 = 4\times81324=4×81 we deduce that aaa divides 444. Thus rrr must be one of ±13,±23,±43\pm\tfrac13,\pm\tfrac23,\pm\tfrac43±31,±32,±34. Since none of these six numbers is a zero of fff, this case does not work.
If b=1b=1b=1 then r=ar=ar=a and f(a)=0f(a)=0f(a)=0. Since 2a5=27a(2−a3)3+81(2−a3)22a^5 = 27a(2-a^3)^3 + 81(2-a^3)^22a5=27a(2−a3)3+81(2−a3)2, we deduce that 333 divides aaa. If 323^232 divided aaa, then 353^535 would divide f(a)+324=324f(a) + 324 = 324f(a)+324=324, which it does not. Thus r=ar = ar=a must be one of ±3,±6,±12\pm3, \pm6,\pm12±3,±6,±12. Since none of these six numbers is a zero of fff, this case does not work either.
There are no nonzero rational roots to these simultaneous equations.
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'Thus we deduce that b3 b^3 b3 divides 27a10 27a^{10} 27a10 ', Can you explain this to me..?? I didnt get this part. Thanks for the solution BTW.. :)
b3b^3b3 is a factor of all the other terms.
@Mark Hennings – This is wonderful! How did you come up with this? Are you some sort of math professor?
@Kevin H – I've been doing Maths for a while. This problem is fiddly, but did not need tough ideas to solve.
@Mark Hennings – MASTER
@Mark Hennings – Thanks Sir! Now i get it
wow that was fabulos can i meet u in fb
'Find x and y so that they are rational'.. but isnt Diophantine Equation about integer solutions?
Thank you for bringing this up I may need to clarify this problem.
You're Welcome.. but this is a good question anyways.. I'm still working on it! :P
Lol, x=y=0x=y=0x=y=0 :P
Lol sorry :D x and y≠0
could you please post the answer to this question,kevin
there are nonzero, I am waiting for the real problem
there are nonzero solutions, please give the real problem
Put x as ky and solve
But that would still mean we have 2 variables.
Also, k will be a rational number to maintain generality, possibly complicating the question.
@Keshav Vats – LOL, hi Keshav.. :P
@Keshav Gupta – :D
what do ypu want exactely, to solve problem ou to do something, like jocking how many solotion do you want wat is the problem please, i need more details for discussing with you i am from morroco nice to know and meet you
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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If both x and y are rational and nonzero, then r=yx is rational and nonzero. Since 6y3=3x3+xy3, we deduce that x=6−3r3, and of course y=rx=r6−3r3. From the first equation we deduce that 2x32x322r5====x2y4+9y5x2r4x4+9r5x5=r4x6+r59x5r4x3+r59x2rx3+9x2=27r(2−r3)3+81(2−r3)2 and hence f(r)=27r10−162r7−81r6+2r5+324r4+324r3−216r−324=0 If we write r=ba where a,b are coprime integers with b≥1, then 27a10−162a7b3−81a6b4+2a5b5+324a4b6+324a3b7−216ab9−324b10=0 Thus we deduce that b3 divides 27a10, and hence b3 divides 27, and hence b=1,3. Moreover a divides 324b10, and so a divides 324.
If b=3 then, since 324=4×81 we deduce that a divides 4. Thus r must be one of ±31,±32,±34. Since none of these six numbers is a zero of f, this case does not work.
If b=1 then r=a and f(a)=0. Since 2a5=27a(2−a3)3+81(2−a3)2, we deduce that 3 divides a. If 32 divided a, then 35 would divide f(a)+324=324, which it does not. Thus r=a must be one of ±3,±6,±12. Since none of these six numbers is a zero of f, this case does not work either.
There are no nonzero rational roots to these simultaneous equations.
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'Thus we deduce that b3 divides 27a10 ', Can you explain this to me..?? I didnt get this part. Thanks for the solution BTW.. :)
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b3 is a factor of all the other terms.
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wow that was fabulos can i meet u in fb
'Find x and y so that they are rational'.. but isnt Diophantine Equation about integer solutions?
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Thank you for bringing this up I may need to clarify this problem.
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You're Welcome.. but this is a good question anyways.. I'm still working on it! :P
Lol, x=y=0 :P
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Lol sorry :D x and y≠0
could you please post the answer to this question,kevin
there are nonzero, I am waiting for the real problem
there are nonzero solutions, please give the real problem
Put x as ky and solve
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But that would still mean we have 2 variables.
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Also, k will be a rational number to maintain generality, possibly complicating the question.
Log in to reply
Log in to reply
what do ypu want exactely, to solve problem ou to do something, like jocking how many solotion do you want wat is the problem please, i need more details for discussing with you i am from morroco nice to know and meet you