Determine the minimum value of a positive integer constant K such that the equation:
X*Y^{2}  Y^{2} + 2*X + Y = K has precisely four solutions in positive integers.
There is a solution for
K=60(x,y)= (20,1) (6,3) (4,4) (7,2) and a fifth (unintended?) (1,58).
I give it as a possible solution because I didn't search exhaustively for K<60, but I'll be surprised if lower solutions show up.
To solution:
f(x,y) is crescent for all (x,y)>(1,0). This means that every solution will have a higher and a lower coordinate than each other solution.
Let be (x1,y1) and (x2,y2) two solutions. It should be that the increment of the value of f(x,y) when we increase the coordinate y from y1 to y2 a value "b", will be compensate by the decrement of the value of f(x,y) when decreases the coordinate x from x1 to x2 a value "a".
Translate into equation:
b(2y1+b)(x11)+b = a(y1+b)^2+2a
[b(2y1+b) here comes from formula of quadratic increment a b value]
Now we can start picking up y1=1 and see if leads to somewhere
For b=1 there is no integer value for x1 in the formula.
For b=2; x11=(11a2)/8
 For a=(6, 14, 22, ...) x1 is an integer. x1=(9, 20, 31,...) and f(x,y)=(27, 60, 93,...)
 Then we obtain (x2,y2) = (x1a, y1+b) = (3,3) (6,3) (9,3)...
For b=3; x11=(18a3)/15
 For a=(1, 6, 11, 16,...); x1 is integer , x1=(2, 8, 14, 20,..) and f(x,y)= (6, 24, 42, 60,..)
 Then we obtain (x3,y3) = (1,4) (2,4) (3,4) (4,4)... concretely for f(x,y)=60, (x,y)=(4,4).
Now we could still doing this, but as we have three values of (x,y) with K=60, we can try to have the four one directly from f(x,y)
Probing for y=5, 6, 7... in f(x,y) leads to (x4,y4)=(2,7)
Note that no other values of (x,y) will be possible for y>=8 and x >1 as then always f(x,y)>60.
Note. But now I see that also there is a solution for x=1, y= 58. Was this intended?
Edited on April 26, 2016, 1:11 pm

Posted by armando
on 20160426 11:43:56 