What is the smallest integer $N>2$, such that $x^5+y^5 = N$ has a rational solution?

Question

I was considering the Fermat-like equation, $$a^k + N b^k = c^k$$

which, for odd power $k$, is equivalent to $$x^k+y^k = N\tag1$$

and rational $x,y$.

Q: For a given $k$, what is the smallest integer $N$ such that $(1)$ has a solution with rational ($\color{red}{not}$ integer) $x,y$?

For $k=3$, and appealing to the theory of elliptic curves, it is $N=6$:

$$\bigl(\tfrac{17}{21}\bigr)^3+\bigl(\tfrac{37}{21}\bigr)^3 = 6$$ $$\bigl(\tfrac{-1805723 }{960540 }\bigr)^3+\bigl(\tfrac{2237723 }{960540 }\bigr)^3 = 6$$

For $k=5$, it seems to be the horribly large $N=68101$:

$$\bigl(\tfrac{15}{2}\bigr)^5+\bigl(\tfrac{17}{2}\bigr)^5 = 68101$$

However, my search radius was very limited, with numerator and denominator $<500$. As the second example for $k=3$ shows, it may not been large enough. So what really is $N$ for $k=5$?

So you explicitly disallow integer solutions? Otherwise you trivially get $2^5+1^5=33$. — Wojowu, Jan 13 '18 at 15:45
@Wojowu: Yes, to be analogous with $N=6$. I made sure to highlight it in red. :) — Tito Piezas III, Jan 13 '18 at 15:51
@Wojowu: If I allow integers, the smallest for *any* $k$ is $N=2$ and deflates the question. — Tito Piezas III, Jan 13 '18 at 15:54
There is a proof by Seiji Tomita at http://www.maroon.dti.ne.jp/fermat/dioph64e.html. Someone who can verify it should post it as an answer. — Ricardo Buring, Jan 14 '18 at 08:41
I currently looking at a problem that may be related, for $k=6$. I’ll look at this next. — Old Peter, Jan 17 '18 at 10:35
@OldPeter: It turns out Tomita has investigated this though. — Tito Piezas III, Jan 17 '18 at 13:53
@RicardoBuring: Thanks for the reference. With remarks, it is enough as a preliminary answer. — Tito Piezas III, Jan 17 '18 at 14:51

Tito Piezas III · Accepted Answer · 2018-01-17T14:55:36.823

Thanks to R. Buring's comment, it turns out A111152 has,

$$(17/21)^3 + (37/21)^3 = 6$$

$$(25/17)^4 + (149/17)^4 = 5906 $$

$$(15/2)^5 + (17/2)^5 = 68101$$

with D. Winter remarking in a 2001 sci.math post that "...it is known 68101 is the smallest n for k = 5".$^\text{his reference?}$

For higher powers $k$, it seems it depends on $k \equiv \text{mod}\, 4$. Seiji Tomita has conjectured that for $m\geq1$,

$$n=\Bigl(\tfrac{2^{4m}-1}{2}\Bigr)^{4m+1}+\Bigl(\tfrac{2^{4m}+1}{2}\Bigr)^{4m+1}$$

$$n=\Bigl(\tfrac{p}{5}\Bigr)^{4m+2}+\Bigl(\tfrac{q}{5}\Bigr)^{4m+2}$$

$$n=\Bigl(\tfrac{2^{4m+2}-1}{2}\Bigr)^{4m+3}+\Bigl(\tfrac{2^{4m+2}+1}{2}\Bigr)^{4m+3}$$

and where $p+q\,i =(2+i)^k$ for appropriate $k\leq4m+2$.

The difficult case is $k=4m$. Tomita found,

$$(25/17)^4 + (149/17)^4\\ (50429/17)^8 + (43975/17)^8\\ (9298423/17)^{12} + (8189146/17)^{12}$$

though the numerators apparently are not as well-behaved as $p,q$.

What is the smallest integer $N>2$, such that $x^5+y^5 = N$ has a rational solution?

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