Hint: Analyse the denominator First ^_^
Let ( x , y ) ∈ N such that ( x , y ) is the solution of
x + 1 1 + y − 1 1 = 6 5 .
Find x + y .
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3 solutions
5/6 = (3+2)/6 =3/6 + 2/6 =1/2 + 1/3 = 1/(1+1) + 1/(4-1)
let, x = 1 and y = 4 so, x+y = 5
1 > 6 5 = > x + 1 ≥ 2 and y − 1 ≥ 2 .
One of x + 1 1 and y − 1 1 is at least half of 6 5 , so either x + 1 ≤ 5 1 2 = 2 . 4 or y − 1 ≤ 5 1 2 = 2 . 4 .
Consider the two cases:
Case 1, x + 1 = 2 = > x = 1 − − − > y − 1 = 3 = > y = 4 .
Case 2, y − 1 = 2 = > y = 3 − − − > x + 1 = 3 = > x = 2 .
In either case, x + y = 5 .
Mr Paul made us all fool by giving wrong hint :P no offence ! just solve the numerator which is x+y on the RHS numerator is 5 as LHS=RHS --> x+y=5