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\begin{center}
\vskip 1cm{\LARGE\bf
Equal Sums of Two Distinct Like Powers
}
\vskip 1cm
\large
Samuel S. Wagstaff, Jr. \\
Center for Education and Research in Information Assurance \\
and Security, and Department of Computer Sciences\\
Purdue University\\
West Lafayette, IN 47907-1398\\
USA \\
\href{mailto:ssw@cerias.purdue.edu}{\tt ssw@cerias.purdue.edu} \\
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\begin{abstract}
We study the equation $a^j+b^j=c^k+d^k$ for positive
integers $a$, $b$, $c$, $d$, and $24$.
According to Guy \cite[Sec.\ D1]{Guy94},
people have searched at least up to $N<10^{25}$ for solutions
in the case $j=5$.
Fermat's Last Theorem says there are no solutions with $d=0$.
Browning \cite{Bro02} used algebraic geometry to show that
primitive solutions with $j>4$ are rare if they exist at all.
We have searched for solutions to the equation
$a^j+b^j=c^k+d^k$ in integers $a$, $b$, $c$, $d$ with $22$.
We found primitive solutions in the three cases $j=3$, $k=4$, 5, and 6,
but none for other values of $21$ might be considered
if one wants to see all solutions, because sometimes
a common factor cannot be canceled.
For example, one solution to $a^3+b^3=c^6+d^6$ is
\begin{equation}
102^3+330^3=12^6+18^6 \label{eq1}
\end{equation}
and the common factor 6 of the four numbers cannot be canceled
since it appears to different powers on the two sides of the equation.
However, this solution comes from a solution to $w^3+x^3=y^3+z^3$.
If $k=\gcd(a,b,c,d)>1$ for a solution to $a^3+b^3=c^6+d^6$,
then this equation may be rewritten as $(kw)^3+(kx)^3=(ke)^6+(kf)^6$
or $k^3w^3+k^3x^3=(k^2e^2)^3+(k^2f^2)^3$ from which $k^3$ may be
canceled to give $w^3+x^3=(ke^2)^3+(kf^2)^3$, which is primitive.
A solution $(k^2a)^3+(k^2b)^3=(kc)^6+(kd)^6$ for some integer $k>1$
is easily derived from the solution $a^3+b^3=c^6+d^6$ by
multiplying by $k^6$, so it is ignored.
For example, $918^3+2970^3=36^6+54^6$ is just the solution
$102^3+330^3=12^6+18^6$ multiplied by $3^6$.
Likewise, we ignore solutions to $a^3+b^3=c^6+d^6$ with $a=c^2$ and $b=d^2$.
Similar reductions apply also whenever $j1$ the probability is near 1,
so we expect many solutions.
When $2/j+2/k=1$ we have a borderline case
where there may be a few solutions or none at all.
This heuristic argument explains why there are many solutions
to $a^j+b^j=c^k+d^k$ when $j=2$: $2/j+2/k>1$
for every integer $k>1$.
When $j=3$ the argument predicts many solutions for $k=4$ and 5,
just a few solutions when $k=6$, and no solution when $k>6$.
It also predicts no solution for $31$ be a real number and $n$ be a positive integer.
Let the measure of the set of all
sequences containing $n$ be $c_1 n^{-1+1/\gamma}$
and the measure of the complement be $1-c_1 n^{-1+1/\gamma}$.
Here $c_1$ and other $c_i$ used later are appropriate positive constants.
In this case, $c_1$ is chosen so that the measure of the set
of all sequences is 1.
Call this measure the $\gamma$ measure.
The phrase ``almost all sequences $\Ac$'' will mean ``for all $\Ac$
except for a set of sequences of $\gamma$ measure 0.''
Let $P_{\gamma}(n)=c_1 n^{-1+1/\gamma}$ be the probability that $n\in\Ac$.
If $\Ac$ is a sequence of positive integers and $x$ is a real number,
let $A(x)$ be the number of $a\in\Ac$ with $a\le x$.
It is easy to see that $A(x)=(1+o(1))c_1\gamma x^{1/\gamma}$
for almost all $\Ac$.
Hence the $n$-th term of $\Ac$ is $(1+o(1))(n/c_1\gamma)^{\gamma}$
for almost all $\Ac$.
Let $2\Ac$ denote the sequence of all sums $a+a'$
with $a\in\Ac$ and $a'\in\Ac$.
Note that if $\gamma=j\ge2$, then, for almost all $\Ac$,
$\Ac$ and $2\Ac$ have growth
rates similar to those of $(n^j)_{n\ge1}$ and $(S_n^{(j)})_{n\ge1}$,
respectively.
Now let $1<\alpha\le\beta$.
Write $P_{\alpha}(n)=c_1 n^{-1+1/\alpha}$ and
$P_{\beta}(n)=c_2 n^{-1+1/\beta}$.
\begin{theorem}
Let $1<\alpha\le\beta$.
If $2/\alpha + 2/\beta < 1$, then for almost all sequences $\Ac$
in $\alpha$ measure and almost all sequences $\Bc$
in $\beta$ measure the intersection $2\Ac\cap2\Bc$ is finite.
But if $2/\alpha + 2/\beta \ge 1$, then for almost all sequences $\Ac$
in $\alpha$ measure and almost all sequences $\Bc$
in $\beta$ measure the intersection $2\Ac\cap2\Bc$ is infinite.
\end{theorem}
The proof is based on that in Erd{\H o}s and Ulam \cite{EU}.
\begin{proof}
We will prove the first statement by showing that the expected
number $E$ of integers $n$ in the intersection is finite.
We have
\begin{align}\nonumber
E & = \sum_{n=1}^{\infty}
\left(\sum_{u+v=n} P_{\alpha}(u)P_{\alpha}(v) \right)
\left(\sum_{u+v=n} P_{\beta}(u)P_{\beta}(v) \right) \\ \nonumber
& = \sum_{n=1}^{\infty}
c_1^2 c_2^2 \left(\sum_{u+v=n} (uv)^{-1+1/\alpha} \right)
\left(\sum_{u+v=n} (uv)^{-1+1/\beta} \right) \\ \nonumber
& < c_3 \sum_{n=1}^{\infty} n^{-1+2/\alpha} n^{-1+2/\beta} \\ \nonumber
& = c_3 \sum_{n=1}^{\infty} n^{-2+2/\alpha+2/\beta} , \nonumber
\end{align}
where the inner sums were estimated by integrals.
For example, the first inner sum is estimated by
$\int_0^n (u(n-u))^{-1+1/\alpha}du=n^{-2+2/\alpha}nc_4=c_4n^{-1+2/\alpha}$.
Since $2/\alpha + 2/\beta < 1$, $E<\infty$ and the intersection
is finite by the Borel-Cantelli lemma.
See Feller \cite{Fell} for the Borel-Cantelli lemma.
Now suppose $2/\alpha + 2/\beta \ge 1$.
We will give the proof for the case $2/\alpha + 2/\beta = 1$.
The case $2/\alpha + 2/\beta > 1$ is similar.
Let $E(x)$ denote the expected number of integers $n\le x$ in the intersection.
We have
\begin{align}\nonumber
E(x) & = \sum_{n=1}^x
\left(\sum_{u+v=n} P_{\alpha}(u)P_{\alpha}(v) \right)
\left(\sum_{u+v=n} P_{\beta}(u)P_{\beta}(v) \right) \\ \nonumber
& = \sum_{n=1}^x
c_1^2 c_2^2 \left(\sum_{u+v=n} (uv)^{-1+1/\alpha} \right)
\left(\sum_{u+v=n} (uv)^{-1+1/\beta} \right) \\ \nonumber
& = (1+o(1)) c_5 \sum_{n=1}^x n^{-2+2/\alpha+2/\beta} \\ \nonumber
& = (1+o(1)) c_5 \sum_{n=1}^x n^{-1} = (1+o(1)) c_5 \log x . \nonumber
\end{align}
Now we use a second moment argument to show that
for almost all $\Ac$ and almost all $\Bc$ the size $f(\Ac,\Bc,x)$
of $2\Ac\cap2\Bc\cap[1,x]$ satisfies
$f(\Ac,\Bc,x)=(1+o(1)) c_5 \log x $ so that
$\lim_{x\rightarrow\infty} f(\Ac,\Bc,x)/E(x) = 1$.
The expected value of $f(\Ac,\Bc,x)$ is $E(x)$, which we just
computed.
Let $E^2(x)$ be the expected value of $f(\Ac,\Bc,x)^2$. Then
$E^2(x)=$
\[
\sum_{n_1=1}^x\sum_{n_2=1}^x
\sum_{u_1+v_1=n_1}
\sum_{u_2+v_2=n_1}
\sum_{u_3+v_3=n_2}
\sum_{u_4+v_4=n_2}
P(u_1,u_2,u_3,u_4,v_1,v_2,v_3,v_4),
\]
where $P(u_1,u_2,u_3,u_4,v_1,v_2,v_3,v_4)$ is the probability
that $u_1$, $v_1$, $u_2$, and $v_2$ are in $\Ac$ and
that $u_3$, $v_3$, $u_4$, and $v_4$ are in $\Bc$.
If these eight integers were distinct, we would have
$P(u_1,u_2,u_3,u_4,v_1,v_2,v_3,v_4)=$
\[
P_{\alpha}(u_1)
P_{\alpha}(v_1)
P_{\alpha}(u_2)
P_{\alpha}(v_2)
P_{\beta}(u_3)
P_{\beta}(v_3)
P_{\beta}(u_4)
P_{\beta}(v_4),
\]
and the sum would be $(E(x))^2$,
but if some integers are repeated, the probability is larger.
Hence, $E^2(x)>(E(x))^2$.
To get the opposite inequality we add terms to $E^2(x)$
to account for possible repeated values.
If the eight integers were distinct, we could pair the sums
\[
\left(\sum_{u_1+v_1=n_1} P_{\alpha}(u_1)P_{\alpha}(v_1) \right)
\left(\sum_{u_3+v_3=n} P_{\beta}(u_3)P_{\beta}(v_3) \right)
\]
and obtain $(1+o(1))c_6\log x$ as in the calculation for $E(x)$.
But if say, $u_3=u_1$, then $v_3=v_1$ and the two sums become
\begin{align}\nonumber
\sum_{u_1+v_1=n_1} P_{\alpha}(u_1)P_{\alpha}(v_1)
P_{\beta}(u_1)P_{\beta}(v_1)
= & \sum_{u_1+v_1=n_1} (u_1v_1)^{-1+1/\alpha} (u_1v_1)^{-1+1/\beta} \\ \nonumber
= & \sum_{u_1+v_1=n_1} (u_1v_1)^{-2+1/\alpha+1/\beta} \\ \nonumber
= & \sum_{u_1+v_1=n_1} (u_1v_1)^{-3/2}. \nonumber
\end{align}
When we approximate this sum by the integral
$
\int_0^{n_1} (u_1(n_1-u_1))^{-3/2}du
$
we get a constant times $n_1^{-2}$, and the sum on $n_1$ is finite.
The sum on $n_2$ pairs the other two sums and gives
$(1+o(1))c_7\log x$.
Likewise, the other added terms are all less than constants times $\log x$.
Therefore, $(E(x))^2 < E^2(x) < (E(x))^2+c_8\log x$.
By the Tchebycheff inequality the $\alpha$ measure of the set of $\Ac$ and
the $\beta$ measure of the set of $\Bc$ for which
\[
|f(\Ac,\Bc,x) - E(x)| > \epsilon\log x
\]
are less than $c_9/\epsilon^2\log x$.
Let $x_k=2^{k(\log k)^2}$. By the Borel-Cantelli lemma
we have
$$\lim_{k\rightarrow\infty} f(\Ac,\Bc,x_k)/E(x_k) = 1.$$
Therefore,
since $f(\Ac,\Bc,x_k)\le f(\Ac,\Bc,x)\le f(\Ac,\Bc,x_{k+1})$
when $x_k < x < x_{k+1}$,
for almost all $\Ac$ and almost all $\Bc$ we have
$\lim_{x\rightarrow\infty} f(\Ac,\Bc,x)/E(x) = 1$.
In the same way,
$\lim_{x\rightarrow\infty} f(\Ac,\Bc,x)/E(x) = 1$ when
$2/\alpha+2/\beta>1$.
Since $E(x)$ is unbounded as $x\rightarrow\infty$,
the intersection $2\Ac\cap2\Bc$ is infinite for
almost all $\Ac$ and almost all $\Bc$ when
$2/\alpha+2/\beta\ge1$.
\end{proof}
Now let $\alpha=j$ and $\beta=k$.
If $(n^j)_{n\ge1}$ and $(n^k)_{n\ge1}$ were typical sequences,
this theorem would predict infinitely many solutions
to $a^j+b^j=c^k+d^k$ when $2/j+2/k\ge1$
and finitely many solutions when $2/j+2/k<1$.
Of course, Lander \cite{Land} found infinitely many solutions
to this equation for all $j$ and $k$.
Thus $(n^j)_{n\ge1}$ and $(n^k)_{n\ge1}$ are special sequences
in the exceptional set of measure 0 when $2/j+2/k<1$.
But since the theorem predicts that there should be no solutions,
perhaps Lander found all solutions in this case.
\section{Acknowledgments}
I thank Bjorn Poonen for suggesting an improvement to
the first heuristic argument.
I thank the anonymous referee for suggestions and references
that improved the manuscript and speeded the program.
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\end{thebibliography}
\bigskip
\hrule
\bigskip
\noindent
2020 \emph{Mathematics Subject Classification}:~Primary 11D41.
Secondary 11Y50.
\noindent
\emph{Keywords}:~equal sums like powers.
\bigskip
\hrule
\bigskip
\noindent
(Concerned with sequences
\seqnum{A003336},
\seqnum{A003347},
\seqnum{A003358}, and
\seqnum{A004999}.)
\bigskip
\hrule
\bigskip
\vspace*{+.1in}
\noindent
Received January 8 2022;
revised version received February 24 2022; March 6 2022.
Published in {\it Journal of Integer Sequences}, March 6 2022.
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Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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