 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Constant from Four Solutions (Posted on 2016-04-24) Determine the minimum value of a positive integer constant K such that the equation:
X*Y2 - Y2 + 2*X + Y = K has precisely four solutions in positive integers.

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Possible solution | Comment 1 of 2
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)(x1-1)+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; x1-1=(11a-2)/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) = (x1-a, y1+b) = (3,3) (6,3) (9,3)...
For b=3; x1-1=(18a-3)/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 2016-04-26 11:43:56 Please log in:
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