1[x][y]=[x]+[y]+{x}+{y}
now LHS is purely integer so {x}={y}=0
[x][y]=[x]+[y]
[x]=[x]/[y]+1
so [x]/[y] is an integer....
[x]([y]-1)=[y]
[x]=[y]/([y]-1) so [y]≠1 and [y]/([y]-1) is an integer
find no of inegral solutions of [x][y] = x + y ................
and find the lines on which the non integral soln lie .............
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4 Answers
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106as x+y is an integer .. so {x}+{y} = 1 or {x}={y} = 0
so the equation can be written as [x][y] = [x]+[y] + 1 or [x][y] = [x]+[y]
106taking [x][y]=[x]+[y] ...
Now, [y] = [x]/([x]-1) .. so [x]-1 is a factor of [x]...
by simple observation, 2,1 and 0,-1 are the only solutions corresponding to [x]/([x]-1)
So, [x]=2 and [x]=0 are the only solutions....
So, the solutions are (0,0) and (2,2)
106taking [x][y] = [x]+[y]+1,
[y] = ([x]+1)/[x]-1)
so the possible pairs are .. 2/0, 3/1(admissible), 4/2(admissible), 5/3, 6/4 ... and 1/-1(admissible), 0/-2(admissible), -1/-3 .. others inadmissible
So, [x]=2, [y]=3 ... [x]=3, [y]=2 ... [x]=0, [y]=-1 ... [x]=-1, [y]=0
Now {x}+{y}=1
