Recently, I’ve been wondering about natural solutions of equation

x^{2} + 7 = 2^{n}^{}

While I’m still working on it, let’s find all natural solutions of similar equation

x^{2} + 1 = 2^{n}

using the same methods of simple algebraic number theory as before. This time my post is rather short.

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We will work in the ring Z[i] of Gaussian integers. It is a ring consisting of elements of form a + bi with a and b being ordinary integers. It is a Unique Factorisation Domain.

Let’s prove that there are no natural numbers (x, n) different from (1,1) satysfying

x^{2} + 1 = 2^{n}

Indeed, this equation is equivalent to

(x + i)(x – i) = (1 + i)^{n}(1 – i)^{n}

Now, what is the Greatest Common Divisor of x+i and x-i? These two numbers are complex conjugates of each other, so it’s easy to see that their GCD must be 2^{k}. But 2^{k} doesn’t divide x+i, for k>0, so we conclude that x+i and x-i are relatively prime. So, using that 1+i and 1-i are primes (for a similar reason I described before), we see that if x and n satisfy our equation then

x+i = (1+i)^{n}

(or … = (1-i)^{n}, but it doesn’t make real difference further). (There are many different ways of ending the reasoning now – you may want to try to do it yourself!) Now, let’s define sequences a_{n} and b_{n} by

(1+i)^{n} = a_{n} + b_{n}i

It’s not hard to see that b_{n} satisfies recurrence relation

b_{n+2}=2b_{n+1} – 2b_{n}

with two first terms equal to 1 and 2.

So, the only odd b_{n} is b_{1}=1. In particular, the only n for which

x+i = (1+i)^{n}

might hold is n=1. And indeed, this gives rise to our only solution: x=1, n=1.

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Let me remind that for finding all solutions to

x^{2} + 7 = 2^{n}

it’s enough to prove that the sequence

b_{1}=1, b_{2}=1, b_{n+2} = b_{n+1} – 2b_{n}

doesn’t have any +-1 for n>13.

## 5 comments

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January 12, 2007 at 10:25 pm

Alon LevyActually, in the x^2 + 1 case, there’s an easier solution. 1+i and 1-i are associate, so you actually have 2^n = (1+i)^2n * (-i)^n. Now x+i and x-i have the same norm, so the fact that they’re powers of the same prime implies that they’re associates. That immediately shows x = (+/-)1.

January 22, 2007 at 7:27 pm

Jakub UrbanowiczHello,

I have an impression that I found a mistake:) GCD(x+i,x-i) is not necessarily equal to 1 (e.g. 1+i divides both 5+i and 5-i). Actually 1+i is the only possible prime common divisor of x+i and x-i because any such divisor would have to divide x+i-(x-i)=2i which prime decomposition is (-i)*(1+i)^2. Anyway, we obtain an equation: (x+i)(x-i) = 2^n = ((1+i)^2n)*(-i)^n. and x+i must be equal to ((1+i)^n)*invertible_element. (This is because x+i and x-i have the same norms, so the exponent of 1+i in x-i and x+i have to be equal). However, I believe that the rest of the reasoning is OK as multiplying by an invertible element does not change the module of imaginary part.

Unfortunately I still don’t know the solution of the 7’s problem.

Anyway, this is a fantastic idea to maintain such a column and to share mathematical and physical ideas with others.

Howgh,

Jakub

January 22, 2007 at 8:04 pm

sirix“I have an impression that I found a mistake”

Jakub, You’re totally right. Thanks for correction.

Also, sorry for your comment being blocked by a spam filter. I only recently learned that such a filter existed and so it took so much time for your comment to appear (I had to manually unblock it).

And, It’s good to know you’re still alive! ;-)

March 10, 2007 at 12:37 pm

John R RamsdenEasier still, in x^2 + 1 = 2^n for n > 0, x must be odd and thus the left side is == 2 mod 8 which forces n = 1 ;-)

March 10, 2007 at 2:50 pm

sirix:-) Ok, Let’s say it was a “proof of a concept” :-)