Prove that for every integer x, there is an integer y such that (y^2-2)/(x^4+1) is an integer.
At the outset, it is trivial to observe that for |x| =1, the quotient
(y^2-2)/2 is always an integer whenever y is even.
It is also trivial to observe that for x =0, the given quotient
reduces to (y^2-2) which is always an integer for any given integer t.
Thus, we will restrict ourselves to the case whenever |x|>=2
By the problem
(y^2-2)/(x^4+1) = p(say), where p is an integer.
Or, y^2 Mod (x^4+1) = 2
By inspection, we find that:
x=2 gives (y,p) = (6,2)
x=3 gives (y,p) = (24, 7)
x=4 gives (y,p) = (60, 14)
This leads us to conjecture that:
y Mod (x^4+1) = x^3-x
Or, y = (x^3-x) +t(x^4+1), where t is an integer......(**)
Substituting the value of y in relationship (**), we
p = (y^2-2)/(x^4+1)
= (x^2-2) + 2t(x^3-x) + t^2(x^4+1)
>= (x^2-2) + (x^3-x)+(x^4+1), whenever t is non negative and x is positive
The said quotient is thus always an integer for non negative t.
When t is negative, let us substitute t=-u,where u is positive,
p = (x^2-2) - 2u(x^3-x) + u^2(x^4+1)
>= (x^4 -2*x^3+x^2+2x -1)
> 0, whenever x>= 2
Thus the case is true for x>2
In a similar vein, it can be shown that p is an integer whenever
Consequently, it follows that there always exist an integer y such that (y^2-2)/(x^4+1) is an integer.
Edited on May 21, 2007, 5:22 am