Fermat's Last Theorem - Variation with arithmetically descending exponents

Question

Are there solution(s) to the following variant of Fermat's Last Theorem in the positive integers?

$$ a^n + b^{n-i} = c^{n-2i} $$

~~I haven't been able to identify any trivial solutions. To my (mathematically untrained) eye, it seems obvious that at least $n > 2i$ must hold.~~

Zev Chonoles noted in early commenting that $n=3$, $i=1$, and $a,b$ arbitrary provides a class of trivial solutions and, indeed, such trivial solutions exist for any $n-2i=1$. So! Are there any solutions to the above equation in the positive integers, where $n-2i>1$?

The above was inspired by this question, and in particular the last portion of Ypnypn's answer to it:

We can also note that $$ \sum_{k=1}^{z-1}{a^k} = {a^z-1\over a-1} < a^z $$

Meaning, even if we added all the exponents of $a$ less than $z$, we still wouldn't reach $a^z$.

Essentially, it pointed to the question, "What could one reach with sums of exponents, if not 'all the way to $a^z$'"? Any corresponding variant of the original linked question is just a subset of the problem posed in that original question, so I went straight to the above form.

There's plenty of trivial solutions with $n=3$, $i=1$, and $a,b$ arbitrary. — Zev Chonoles, Sep 03 '15 at 16:26
@ZevChonoles Ha! Indeed. Are there trivial solutions any time $n-2i=1$? — hBy2Py, Sep 03 '15 at 16:29
Yup, that seems to be the case. Of course, it's your prerogative to disallow such cases. — Zev Chonoles, Sep 03 '15 at 16:30
@ZevChonoles Yep - this is what tends to happen when I post on Math.SE -- it takes several iterations to iron out the trivialities that actual mathematicians see right away. — hBy2Py, Sep 03 '15 at 16:32

score 3 · Answer 1 · answered Sep 03 '15 at 16:51

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Well, the simplest one I could find in a few minutes was $100^4 + 500^3 = 15000^2$.

answered Sep 03 '15 at 16:51

John Brevik

1,782

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$1^4+2^3=3^2$ is simpler. – Julián Aguirre Sep 03 '15 at 16:56
Ha!! Got me. Though I did see that one and discarded it. – John Brevik Sep 03 '15 at 16:58
Another example: $n=6$, $i=2$, $3^6+6^4=45^2$. – Julián Aguirre Sep 03 '15 at 17:07
John or @JuliánAguirre, would you be willing to lay out your thought process in finding these? (If there was one, that is... as opposed to just fiddling/free-searching.) – hBy2Py Sep 03 '15 at 17:07
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It was a search with Mathematica. – Julián Aguirre Sep 03 '15 at 17:12
@JuliánAguirre Ahh, brute force. :-) – hBy2Py Sep 03 '15 at 17:19
I just noticed that 100+125=225 and figured I could tweak a 4th power in there. Not too helpful either I guess. – John Brevik Sep 03 '15 at 17:49
Julián's uses the well known identity of sqares $3^4( 3^2+4^2)=3^4(5^2)$ Perhaps such an ansatz allows a structure for more solutions... – Gottfried Helms Sep 04 '15 at 10:01

score 1 · Accepted Answer · answered Sep 04 '15 at 07:33

Assume $\gcd(n,2i)=1$. Given any $x$, $y$, let $x^n+y^{n-i}=w$. Then for any $r$ we have $$x^nw^{rn(n-i)}+y^{n-i}w^{rn(n-i)}=w^{1+rn(n-i)}$$ which is to say $$(xw^{r(n-i)})^n+(yw^{rn})^{n-i}=w^{1+rn(n-i)}$$ Since $n$ (and, thus, $n-i$) is prime to $n-2i$, we can choose $r$ such that $1+rn(n-i)=(n-2i)s$ for some integer $s$, and so we have a solution of $a^n+b^{n-i}=c^{n-2i}$.

Much more interesting would be to find solutions with $a,b,c$ pairwise coprime. I suspect that all the work that has been done on the general $a^r+b^s=c^t$ since Wiles-Taylor has already settled this problem.

Fermat's Last Theorem - Variation with arithmetically descending exponents

2 Answers2