Diophantine Equation
Find the number of ordered pairs ( x , y ) of positive integers such that x y − 1 2 y = 2 3 5 + 1 3 x
The answer is 4.
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4 solutions
Yea, same way :) Short and sweet
yeah thats the way I solved....
This solution is great!!!!!!
The equation can be written as ( y − 1 3 ) x = 1 2 y + 2 3 5 That is x = y − 1 3 1 2 y + 2 3 5 .
The RHS fraction can be written as y − 1 3 1 2 y − 1 5 6 + 1 5 6 + 2 3 5 and splitting the fraction, the equation becomes x = 1 2 + y − 1 3 3 9 1 .
Since the problem asks for positive integer solutions, y − 1 3 must be a positive divisor of 3 9 1 or a negative divisor such that ∣ y − 1 3 3 9 1 ∣ ⩽ 1 2 (in this case, this inequality has no integer solutions), and since 3 9 1 is 1 7 × 2 3 , our solutions will be:
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from y − 1 3 = 1 7 ,we have ( 3 5 , 3 0 ) ;
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from y − 1 3 = 2 3 , we have ( 2 9 , 3 6 ) ;
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from y − 1 3 = 3 9 1 , we have ( 1 3 , 4 0 4 ) ;
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from y − 1 3 = 1 , we have ( 4 0 3 , 1 4 )
for a total of 4 solutions.
x>0, not x>=0, so | 391 \ (y-13) | < 12. (btw, how to add LaTex? I know about the language but don't know how to insert it here. on Mathematics Stack Exchange, e.g., we use $[LaTex text]$).
good and beautiful!!
The expression can be rewritten as (x - 12)(x - 13) = 391 suppose 391 = 17(23), it implies that (x, y) has four solutions.
Please change the question to positive integer ordered pairs because if we include negative integers, we will have 8 ordered pairs.
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You're correct ....my mistake...Sorry for the inconvenience........
The equation can be rewritten as: ( x − 1 2 ) ( y − 1 3 ) = 3 9 1 = 1 × 3 9 1 = 1 7 × 2 3 For each factorization, we have 2 ordered solution. Hence there are total 4 solutions.