Some Problem Solving Techniques

Changing the question . Find all k such that the diophantine equation x²+y²+1 = kxy has solution in positive integers.

xy 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

that'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

Great work , thanks .

oops, yes ricky u r right.. mine wa smistake

and dont call me sir...

i am neither that old nor that worthy of the title