m 1 − n 1 = 6 1 . How many integral ordered pairs ( m , n ) are there?
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Simon's favorite factorizing trick is the best approach to solving such Diophantine equations.
(Taken from Eamon's comment from a deleted solution)
The entire list is ( 5 , 3 0 ) ( 4 , 1 2 ) ( 3 , 6 ) ( 2 , 3 ) ( − 3 , − 2 ) ( − 6 , − 3 ) ( − 1 2 , − 4 ) ( − 3 0 , − 5 ) ( 7 , 4 2 ) ( 8 , − 2 4 ) ( 9 , − 1 8 ) ( 1 0 , − 1 5 ) ( 1 2 , − 1 2 ) ( 1 5 , − 1 0 ) ( 1 8 , − 9 ) ( 2 4 , − 8 ) ( 4 2 , − 7 )
It should be written that how many integral ordered pairs are there. If you don't mention that then the answer is infinite.
I believe the answer should be 9 and I am going to mention the possible solutions of this problem, let me know if I missed anything ( − 6 , − 3 ) ( 3 , 6 ) ( 7 , − 4 2 ) ( 8 , − 2 4 ) ( 9 , − 1 8 ) ( 1 2 , − 1 2 ) ( 1 8 , − 9 ) ( 2 4 , − 8 ) ( 4 2 , − 7 )
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This is my entire list: ( 5 , 3 0 ) ( 4 , 1 2 ) ( 3 , 6 ) ( 2 , 3 ) ( − 3 , − 2 ) ( − 6 , − 3 ) ( − 1 2 , − 4 ) ( − 3 0 , − 5 ) ( 7 , 4 2 ) ( 8 , − 2 4 ) ( 9 , − 1 8 ) ( 1 0 , − 1 5 ) ( 1 2 , − 1 2 ) ( 1 5 , − 1 0 ) ( 1 8 , − 9 ) ( 2 4 , − 8 ) ( 4 2 , − 7 )
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The equation can be written as 6 n − 6 m = m n
m n − 6 n + 6 m − 3 6 = − 3 6
( m − 6 ) ( n + 6 ) = − 3 6
Now 3 6 = 2 2 × 3 2 so it has ( 2 + 1 ) ( 2 + 1 ) = 9 factors, however, since − 3 6 must have at least 1 negative factor, then it has 9 × 2 = 1 8 factors (including negative factors).
With values m = 0 and n = 0 :
( 0 − 6 ) ( 0 + 6 ) = − 3 6
But m , n = 0 since it would make the original expression undefined.
So in total there are 1 8 − 1 = 1 7 integral solutions for ordered pairs ( m , n ) .