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It seems if base number $a$ is a natural number and the exponent $n$ is an odd number greater than or equal to $3$, then:

$f(a, n) =\displaystyle\sum_{i=1}^{a^{n-(n+1)/2}}{(2ai-a)}=a^n$

Such as $f(3, 7) = 2187 = 3 ^ 7$

How does this sum work? How about if the exponent is an even number?

Source: "Linearization" of Beal problem

Debashish
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miket
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4 Answers4

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If you take out $a$ as a common factor, what is left is the sum of all odd numbers from $1$ to $2b-1$, where $b=a^{n-(n+1)/2}$. It is well known that this sum is $b^2$, so the total is $$f(a,n)=ab^2=a\Bigl(a^{n-(n+1)/2}\Bigr)^2=a\times a^{2n-n-1}=a^n\ .$$

If $n$ is even then $n-\frac{n+1}{2}$ is not an integer, so $a^{n-(n+1)/2}$ is not an integer (usually: $a$ might be square). So the upper limit of summation would have to be $$\Bigl\lfloor a^{n-(n+1)/2}\Bigr\rfloor\ ;$$ the same argument gives $$f(a,n)=a\Bigl\lfloor a^{n-(n+1)/2}\Bigr\rfloor^2\ ,$$ but it may not be easy to simplify this.

David
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Not too hard to show if you break apart the sum.

$$f(a, n) =\sum_{i=1}^{a^{n-\frac{n+1}{2}}}{(2ai-a)}=2a\sum_{i=1}^{a^{n-\frac{n+1}{2}}}{i}-a\sum_{i=1}^{a^{n-\frac{n+1}{2}}}{1}$$

Now, we just need the general forms for these sums.

$$\sum_{i=1}^n i=\frac{n(n+1)}{2}$$ $$\sum_{i=1}^n 1=n$$

Can you work it from here?

Silynn
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Nice to know that someone is interested to my long work on the way to Euler, Fermat & Beal.

Now is just question to play (so shift to adjust) with 2 linear segments to find the solution.

Ciao Stefano

http://mymathforum.com/number-theory/44983-linearization-beal-problem.html#post199250

stefano
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Sorry i forgot to answer for Even:

you have just to reduce at n=2 remembering that:

$$ A^2 = \sum_{m =1}^{A}{(2m-1)} $$

So you can now play with lines (segments of course) insthead of powers and see what Euler, Fermat and Beal ask you... The answer to the problem require just to play with segments, cutting till B+1 and looking what must/can be done to arrange the two segments from B+1 to C.

Some has a solution: 3^5 = 3^3 + 6^3

since you can arrange 2B^2-B and 2A^2-A with B=6 ; A=3 ; C base=3 (too) in this easy case.

some no: C^3<>A^3+B^3 because A,B,C coprimes so no way to shift (in the integers) 2B^2-B to fit an integer C^n

Thansk Ciao Stefano

stefano
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