Finding common ground
Find number of pairs of integers m , n such that 2 m n − 5 m + n = 5 5 .
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1 solution
Can u just help me with factorizing the above expression step by step
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The best approach for solving a Diophantine equation of this type is to make use of a variation of Simon's Favorite Factoring Trick .
Focusing on the left-hand side of the given equation, we can factor it as
2 m ( n − 2 5 ) + 1 ∗ ( n − 2 5 ) + 2 5 = ( 2 m + 1 ) ( n − 2 5 ) + 2 5 = 2 1 ( 2 m + 1 ) ( 2 n − 5 ) .
Nice solution
The given equation can be rewritten as
( 2 m + 1 ) ( n − 2 5 ) + 2 5 = 5 5 ⟹ ( 2 m + 1 ) ( 2 n − 5 ) = 1 0 5 .
Now 1 0 5 = 3 ∗ 5 ∗ 7 has 2 3 = 8 positive divisors and thus 2 ∗ 8 = 1 6 integer divisors in total. Since each of these divisors is odd, and since each of ( 2 m + 1 ) and ( 2 n − 5 ) is odd, we can in turn assign each of the 1 6 divisor "pairs", (i.e., pairs of integers whose product is 1 0 5 ), to ( 2 m + 1 ) and ( 2 n − 5 ) to obtain unique integer pairs ( m , n ) that satisfy the given equation. Thus the desired answer is 1 6 .