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Old 2017-02-09, 04:08   #1
carpetpool's Avatar
Nov 2016

23·41 Posts
Post Solutions to a^2-ab+b^2 = 3^n

The basic algebra rules for factoring sums and differences of powers are here. For a^n+-b^n when n is prime, (a^n+-b^n)/(a+-b) = 0 or 1 (mod n). If (a^n+-b^n)/(a+-b) = 0 (mod n), let n^k be the highest power of n dividing (a^n+-b^n)/(a+-b). Then (a^n+-b^n)/((a+-b)*n^k) = 1 (mod n). Would it ever be the case that (a, b) > 1, k > 1, (a^n+-b^n)/((a+-b)*n^k) = 1?

For n = 3, we are looking to find

Solutions to a^2+ab+b^2 = 3^n (I messed up in the question title btw, this is what I meant to ask.)

The only known solutions with integers are

1^2+1*1+1^2 = 3^1
2^2-2*1+1^2 = 3^1

It is probably the case that no more solutions with (a, b) > 0 exists other than these two. If you know of one, or have a proof that no solutions to a^2+ab+b^2 = 3^n exist except (1, 1) and (-2, 1), please post arguments here.

Last fiddled with by carpetpool on 2017-02-09 at 04:47
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Old 2017-02-09, 06:12   #2
Batalov's Avatar
Mar 2008

59×163 Posts

Originally Posted by carpetpool View Post
... except (1, 1) and (-2, 1)
How 'bout (2, -1)?
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Old 2017-02-09, 06:41   #3
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Jun 2011

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Tip: you will first note that the expressions are symmetric in a and b. Then, the two are equivalent, as \(a^2+ab+b^2\) is \((a+b)^2-ab\) therefore if you substitute \(a=c+b\) you will have \((c+b)^2-(c+b)b+b^2=c^2+2cb+b^2-cb-b^2+b^2=c^2+cb+b^2\). So, if there is a solution for plus side, then there is a solution for the minus side if you increase \(a\) with \(b\).

Given that, think about the modularity of a and b to 3. The plus expression can only be a multiple of 3 if a and b are both either 0, 1, or 2, (mod 3). For (0, 1), (0, 2), or (1, 2) you always get \((a+b)^2-ab=1\) (mod 3). (others are symmetrical). For the minus expression, the only valid groups are (0, 0) and (1, 2) (with symmetry) otherwise again, all the other combinations result in 1 (mod 3).

Now, for both cases, if (a, b)=(0, 0) (mod 3), then you can divide by 3 on both sides of your equation, and you get a smaller solution (see infinite descent method).

So, you only have to study the plus cases when (a, b)=(1, 1), or (2, 2), and the minus case when (a, b)=(1, 2). Due to plus/minus equivalence, you have only to study a single case. Now, tell us the solutions... [Hint 2: the left side is always 3 (mod 9) therefore no solution with n>=2 can exist ]

Last fiddled with by LaurV on 2017-02-09 at 07:07
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