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man111 (54)

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(1) The no. of positive integral solution of the equation


7x2-2xy+3y2 = 27 


ans = 1, 2 ,0 ,None of these.


(2) find the no. of positive integral solution


x3-y2 = 2

    
Avinash_Bhat (655)

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I think the answer for the 1st question is 2




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rahul1993 (438)

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for the 2nd question


x3=y2+2


x=(y2+2)1/3


now the only solution i could think of was x=3,y=5

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kaymant (1542)

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Yes, Rahul.. you have guess it right. The only positive integral solution for the equation x3 - y2 = 2 is (3 , 5). This equation is an example of Mordell's equation which is an elliptic curve:


                                                          y2 = x3  + n


for an integer n. Till 2006, the general soultion for the above equation had not been found. Follow this link to know a bit more about (if you are interseted):


http://www.lrz-muenchen.de/~hr/numb/mordell.html

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kaymant (1542)

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For the first part, the answer is again 2: the roots are (0, 3), (0,-3). I am trying to find out whether we could find it out using some rigorous means
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rahul1993 (438)

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i am not sure of the answer. so please correct if wrong



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hsbhatt (5581)

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The first one looks like case-work to me.


Since only positive integral solutions are asked for, the pairs (0,3) and (0,-3) are not counted.


We can rewrite the equation as


(x-y)^2 + 2(3x^2+y^2) = 27


x=y gives no integer solutions.


Since the second term is even and is atleast 14 (since x and y are atleast 1 and x, y are unequal), (x-y)2 must be odd and less than or equal to 9


Hence the only possibilites are


1\\ x-y = \pm 1 \\<br/>3x^2 + y^2 = 13


 


2\\ x-y = \pm 3 \\<br/>3x^2 + y^2 = 9


Case 2 gets straightaway ruled out as y must be divisible by 3 and the least such +ve integer is 3 which means x = 0 which is not possible as x is also positive.


Case 1: y can be 1, 2 or 3.


 y =1, x =2 is the only integral solution obtained


Hence (2,1) is the only solution among positive integers


Time wounds all heels
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rahul1993 (438)

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another way of solving


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