Today my teacher told me about a following math problem: find all natural numbers n and x such that equality

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

holds (I will call this equality (1)). Everyone who competed in various mathematical olimpiads have seen lots of similar problems. Why this one is particularly interesting? My teacher was told about it by one of his highschool pupils who just came back from a “Zwardon math camp” (link in polish) – a place where best-in-mathematics polish highschool students come to train before International Mathematical Olympiad. All this __very __clever guys tried to work out the solution but they failed.

So, it might be just that there is no “elementary” solution. However, my teacher told me about what he came up with using some simple algebraic number theory (though, there is no solution yet – I hope somebody reading this can provide me with complete solution, this or another way).

*

“Biggest” solution of (1) known to me is x=181, n=15. I was told that someone checked that for 15 < n < 60 there are no solutions.

Consider following sequence defined by induction:

b_{1} = b_{2} = 1, b_{n}=b_{n-1} – 2b_{n-2}.

* Conjecture: *(1) has a solution for given k <=> b

_{k-2}= 1 or b

_{k-2}= -1

I’ll sketch a proof of “<=” part, and we’ll see a problem that doesn’t allow us to be sure of “=>” part. However, for all known solutions it is true that b_{k-2}=+-1.

Also, even taking conjecture granted, I can’t prove that there are no solutions to (1) with n>15. However, my teacher came up with some conjectures concering sequence b which solve the problem and which were numerically verified in some cases (most important, that b_{k} (mod 2^{l}) is periodic from some point (this is obvious) with a smallest period equal to 2^{l-2}, for all l).

* Proof of “<=” part of Conjecture: *We’ll work in a field Q(sqrt(-7)).

…and due to time being 3:00AM and me going to Warsaw at 9:00AM it’s all for now :-). Usually after getting back to Warsaw it takes me few days to get used to working, so I probably won’t post the rest for a couple of days.

**24.01.2007: Well, I wrote second part – “Elementary arithmetical question, part II” and in was published for a couple of weeks but it was accidentally deleted and I don’t have a copy on my computer. Sorry for that – however, if I finally come up with a complete solution, I’ll write everything from a beginning.**

But, that’s even better! Now, You can find Yourself proof of the conjecture (it’s rather simple, one uses basic properties of ring of integers of Q(sqrt(-7)) AND You have more time to provide me with an answer concerning equation (1) :-). Have a good night!

P.S. To obtain subscript I used HTML tag <sub> – is it ok? Or maybe not everyone is seeing this properly and I should stick with less readable but safer b_k ?

## 3 comments

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January 13, 2007 at 4:31 pm

Alon LevyTo prove the conjecture, work with respect to the integral basis {1, (1+SQRT(-7))/2}. You eventually get that (x(+/-)SQRT(-7))/2 is an kth power of (1+SQRT(-7))/2 iff b(k) = (+/-)1.

There’s a full proof of the theorem here, but I remember having seen easier ones in books I’ll get as soon as my math library reopens, which should be on Monday or Tuesday.

January 15, 2007 at 4:23 pm

sirixI suspect there indeed should exist simpler (shorter) proof, based on investigation of b_n sequence mod 2^n, or maybe in 2-adics.

However, big thanks for reference.

January 12, 2011 at 2:57 am

wpolscemamymocneseoHi….thanks for this good information.I’m so happy because it’s very useful for my thesis research.I hope you will keep updating your content constantly as you have one dedicated reader here :)