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That is why I improved the NSolve command to only solve for Real numbers ( no imaginary).

You should see that only x = 59 and n = 6 are integers.

Since we know that there were only 4 possible solutions because there were only 4 pairs of divisors that when multiplied equal 615 we are almost done. Do you follow up to here?

There will be one more trick.

I am sorry I inadvertently used k instead of n, let us go back to n and x as our variables. Check the whole calculation out now:

```
NSolve[{(2^n + x) == 615, (2^n - x) == 1}, {x, n}, Reals, WorkingPrecision -> 50]
NSolve[{(2^n + x) == 205, (2^n - x) == 3}, {x, n}, Reals, WorkingPrecision -> 50]
NSolve[{(2^n + x) == 123, (2^n - x) == 5}, {x, n}, Reals, WorkingPrecision -> 50]
NSolve[{(2^n + x) == 41, (2^n - x) == 15}, {x, n}, Reals, WorkingPrecision -> 50]
```

Solve for k and x.

`NSolve[{(2^k + x) == 615, (2^k - x) == 1}, {x, k}, Reals, WorkingPrecision -> 50]`

What do you conclude?

Starting with the first one:

It is reasonable to say,

You got 4 pairs of numbers?

You should have {1, 3, 5, 15, 41, 123, 205, 615}.

Now pair them off two at a time to make a product of 615. For instance 1 x 615 = 615.

We will do it step by step and then it will get clear.

When you set the equation up as 2^n - x^2 = 615 you now factor 615

`Divisors[615]`

Hi;

Try this piece of code:

`Factor[2^(2 n) - x^2]`

It takes some math and some code to do that, M can do both.

Hi;

That is the only one there is.

Hi;

Is that e^(2y) in (y/4)+((e2y-e-2y)/16)?

What were you trying to do with it?

Agreed, but that is what we are discussing.

Nor would any sane person, the only thing that would is what I call a zombie.