# Some Problem Solving Techniques

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 .

• Shubhodip ·

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

• Shubhodip ·

oh.. i hvn tried... just changed to feel confident.. now its my q..not urs :P

• Dr.House ·

xy divides x2+y2+1

so x divides x2+y2+1 and y divides x2+y2+1

xâ‰¡0 mod (x2+y2+1) and yâ‰¡0 mod (x2+y2+1)

this implies xâ‰¡0 mod (y2+1) and yâ‰¡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)â‰¡(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

• Ricky ·

Sir , don't you think that it is not obvious , that line no. 4 implies line no. 5 ?

• Shubhodip ·

that's an application of Infinite decent...

Let us consider the case when xâ‰ 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

• Ricky ·

Great work , thanks .

• Dr.House ·

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

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