x

^{2}+ 7y

^{2}= 2

^{3n+3}

has nonzero integer solutions for x and y, for n = 0, or any positive integer.

According to the MathWorld website, check out eqn. 22 on this page, Euler had a similar result with an "unpublished proof":

x

^{2}+ 7y

^{2}= 2

^{n}

^{}

But also according to MathWorld, Ramanujan has a result called Ramanujan's square equation:

x

^{2}+ 7 = 2

^{n}

Which only has solutions for n = 3, 5, 7 and 15. That is listed on the first page I referenced on MathWorld as eqn. 21, but also has its own page.

The Binary Quadratic iterator here, or BQD Iterator for short is:

u

^{2}+ 7v

^{2}= F

means that

(u-7v)

^{2}+ 7(u+v)

^{2}= 2

^{3}*F

The great thing then is that we can actually look inside this thing, and see why it behaves as it does.

The Ramanujan one is actually a little more interesting to me at first because it requires that v = 1 or -1.

Here is the start that gives F = 2

^{3}:

1

^{2}+ 7*1

^{2}= 2

^{3}

So you have u = v = 1, or u = v = -1, is possible I'll note as well, so of course with the iterator those will divide off!

So the next value from the iterator is:

(-6)

^{2}+ 7*(2)

^{2}= (2

^{3})2

^{3}

And of course 4 divides off to give: (-3)

^{2}+ 7 = (2)2

^{3}

And the mystery is solved. Doing further iterations is tedious and boring to me, but I'm guessing if I keep going the 7 would be left bare for exactly as many results as will give Ramanujan's finding, while have u and v must be odd, will always mean their sum gives at

*least*a factor of 4 that divides off.

That explains Euler's result as well, as with each iteration you're multiplying by 8, but dividing 4 off automatically, so net result is multiplying by 2. Wow.

That is kind of interesting, I wonder if Euler or Ramanujan knew the why of these results. I have a mathematical tool that lets me peer inside easily.

With great mathematicians you can never be sure. They may have had this tool and didn't see any reason to tell it, or they did and that's buried somewhere or lost. So I should say, I've seen no record that they had it.

But now we can extend beyond these results, like the next one is:

x

^{2}+ 15y

^{2}= 2

^{4n+4}

has nonzero integer solutions for x and y, for n = 0, or any positive integer.

And we can adjust it a bit immediately by dividing off 4 to:

x

^{2}+ 15y

^{2}= 2

^{2n+4}

has nonzero integer solutions for x and y, for n = 0, or any positive integer, as we know that 4 will divide off.

That gives us a more general result with 2

^{c}where c is a positive integer greater than 1:

x

^{2}+ (2

^{c}- 1)y

^{2}= 2

^{(c-2)n+c}

when c is 2 or greater, has nonzero integer solutions for x and y, for n = 0, or any positive integer.

So we have a general path for Ramanujan type results, as there will always be at least one bare iteration which is:

x

^{2}+ (2

^{c}- 1) = 2

^{(c-2)+c}

Which is of course: x

^{2}+ (2

^{c}- 1) = 2

^{2c-2}

James Harris