sqare side x having area xx?

Question

What is the smallest value of y if:

9xx +1 = yy (y cannot be equal 1)

Answer 1

9x^2 +1 = y^2
1 = y^2-9x^2 = (y+3x)(y-3x)

for integer solutions y+ 3x = 1 or -1 and y - 3x = 1 or -1

y = 1 x =0 or y= -1 x = 0

for non integer it can be anything say y+ 3x = z and y- 3x = 1/z

add to get y = (z+1/z)/2 and subtract to get x = (z-1/z)//6

Answer 2

9x^2 + 1 = y^2

x is a length as x^2 is an area, so x has to be a positive number.

As x approaches 0, y would approach 1.

Eg. x=0.01
9(0.01)^2 + 1 = y^2
9(0.1) + 1 = y^2
y^2 = 1.9
y = SQRT 1.9
y = 1.38

The smaller the value of x, the closer y would be to 1.
It will however not be 1 because the area (x^2) cannot be zero.

Answer 3

It's unclear what your question is supposed to be asking.

If x and y are supposed to be whole numbers greater than zero, then there is no possible value for y.

9xx = (3x)*(3x), which is a perfect square.
If 9xx is a perfect square and x>0, then 9xx+1 can't be a perfect square.

Answer 4

I think you mean y^2 = x^2+1 {y not = 1}
I assume x>0, since x is the side of a square.

So y =+/- sqrt(9x^2 +1)
As x --> 0 y --> +1 and -1
Hence -1 is minimum value . [Note -1 not = 1].

One could also argue that as x--> = infinity, y --> + infinity and - infinity. So - infinity is minimum value.

Take your pick.