21Changing the question . Find all k such that the diophantine equation x2+y2+1 = kxy has solution in positive integers.
Let " x " and " y " be positive integers such that " xy " divides " x 2 + y 2 + 1 " . Prove that -
x 2 + y 2 + 1xy = 3 .
I will add some more if I see that people are taking interest in this .
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7 Answers
2121oh.. i hvn tried... just changed to feel confident.. now its my q..not urs :P
39xy divides x2+y2+1
so x divides x2+y2+1 and y divides x2+y2+1
x≡0 mod (x2+y2+1) and y≡0 mod (x2+y2+1)
this implies x≡0 mod (y2+1) and y≡0 mod (x2+1) dince its obvious that x2 is divisible by x and y2 is divisible by y .
adding these 2 gives us (x+y)≡(x+y)2+2-2xy
this means (x+y) should divide 2-2xy
this means (x+y) should divide 2(1-xy)
since x,y are positive integers, the above condition holds only when x=y=1
therefore (x2+y2+1)/xy=3
1Sir , don't you think that it is not obvious , that line no. 4 implies line no. 5 ?
21that's an application of Infinite decent...
Let us consider the case when x≠y.WLOG x>y
Let x1, y1, n1 be a solution to the diophantine equation x2+y2+1 = nxy.
This gives us (y1n1-x1)2+ y12+1 = (y1n1-x1)y1n1 hence (y1n1-x1),y1,n1 is also a solution..
we have y12+1= x1(y1n1-x1) > 0, implying (y1n1-x1)>0
since x1>y1 , x12>y12+1= x1(y1n1-x1) we have x1 > (y1n1-x1)
Therefore one can form a decreasing sequence of positive integers..but no such sequence exists!
so x=y. This gives x2(n-2) = 1 hence we must have n = 3
39oops, yes ricky u r right.. mine wa smistake
and dont call me sir...
i am neither that old nor that worthy of the title