Diophantine Equation
Given that x and y are integers that satisfy the following equation
x 5 − y 7 = 2 ,
find the sum of all possible values of y .
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1 solution
Did exactly the same each and every step
Can you please briefly describe how did you check for all cases? Also, how do you know the list is exhaustive? Thanks - SK
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you first map 5 - 2x to 5, then -5, then 7, then -7 and the second operand is the 7, -7, then 5 and -5.
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You have described four such cases, but clearly total tuples are seven in number. Even if I then map 5-2x to 10 and 2y+7 to 7/2 and so on, how can I show that the list is exhaustive?
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@Saurabh Kataria – Hi, I have edited my solution.
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@Victor Loh – I now understand the solution, thanks for editing. I also got why the list is exhaustive.
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@Saurabh Kataria – You can check out Simon's favorite factoring trick....
@Saurabh Kataria – Oh, I think x, y are integers, and we have 5 - 2x,and 2y + 7, so that 5 - 2x and 2y + 7 are integers too. Then use permutation to list all case
Hi @Saurabh Kataria , since x and y are both integers, 5 − 2 x and 2 y + 7 are both integers. Also, 3 5 = 1 × 3 5 = 5 × 7 = 7 × 5 = 3 5 × 1 and 3 5 = ( − 1 ) × ( − 3 5 ) = ( − 5 ) × ( − 7 ) = ( − 7 ) × ( − 5 ) = ( − 3 5 ) × ( − 1 ) . However we have to reject ( x , y ) = ( 0 , 0 ) . Hope this helps.
Can you please explain why you chose to multiply both sides by 2xy rather than xy or any other multiple of xy.
Multiplying both sides of the equation by 2 x y , we have 1 0 y − 1 4 x = 4 x y ⟹ ( 5 − 2 x ) ( 2 y + 7 ) = 3 5 = 1 × 3 5 = 5 × 7 = 7 × 5 = 3 5 × 1 and 3 5 = ( − 1 ) × ( − 3 5 ) = ( − 5 ) × ( − 7 ) = ( − 7 ) × ( − 5 ) = ( − 3 5 ) × ( − 1 ) . Checking all cases, we have ( x , y ) = ( 2 , 1 4 ) , ( − 1 , − 1 ) , ( − 1 5 , − 3 ) , ( 3 , − 2 1 ) , ( 5 , − 7 ) , ( 6 , − 6 ) , ( 2 0 , − 4 ) ( 0 , 0 ) must be rejected. So the sum of all possible values of y is − 2 8 .