With Fermat's Last Theorem now proven (not by Fermat!) and formalized (not (yet) by Kevin Buzzard's project! see also this post), it may look like we are at a risk of running out of questions concerning sums of like powers, i.e., expressions of the form $x_1^p+\cdots +x_k^p$ for some $k$ and $p$.

At least, recent developments motivated me to look into such problems, and it turns out that, fortunately (?), there are still many things we don't understand at all. This post is a kind of tour of the numerous open problems that remain about sums of like powers.

Disclaimer: I am not a mathematician. This post talks about a topic well-suited to recreational mathematics in the sense that the statements of the results are elementary (at least I hope I understood them correctly), but the proofs are much more difficult and I don't claim to have the necessary prerequisites. I'm just a curious onlooker, but I may of course be wrong about some points.

Our starting point is Fermat's Last Theorem (FLT). I will give its statement below in case you are not familiar with it (but in this case you should also look at the Wikipedia page to learn about the historical story involving Fermat's margin notes!). FLT shows that there are no positive integer solutions to the following equation for $p>2$:

By contrast, there are infinitely many solutions for $p=2$, which are called Pythagorean triples (they are the triples of integers that can be the length of the sides of a right triangle). This post is divided in four sections covering the following generalizations:

• allowing more terms on one side of the equation, i.e., $x_1^p+\cdots +x_k^p=y^p$
• allowing more terms on either side, i.e., $x_1^p+\cdots +x_k^p=y_1^p+\cdots +y_l^p$
• not forcing the right-hand side (RHS) to be a $p$-th power but in fact aiming at achieving all integers as a sum of $p$-th powers $x_1^p+\cdots +x_k^p$
• looking more generally at the number of ways in which we can achieve each integer as a sum of $p$-th powers

Equations $x_1^p+\cdots +x_k^p=y^p$

For arbitrary powers $p$, how many positive integers to the power $p$ do you need to sum to get another positive integer to the power $p$? Of course we exclude trivial solutions by imposing $k>1$ and requiring that the numbers are strictly positive. FLT then tells you that, for $p>2$, two numbers are not enough, so you need at least 3 numbers, i.e., $k\ge 3$.

Euler conjectured that, in fact, you need at least $p$ numbers, i.e., $k\ge p$. That is, you need to sum three cubes to obtain a cube, four fourth-powers to obtain a fourth-power, and so on. And indeed, for cubes ($p=3$), we do know by FLT that we must have $k\ge 3$. But it turns out that Euler's conjecture is wrong: I'll review this below following Wikipedia's summary. The first refutation came when a computer search in 1966 by Lander and Parkin on the CDC 6600 led to a two-sentence paper giving the identity:

Thus, summing four fifth-powers (i.e., $k=5-1$) is sufficient to get a fifth-power (i.e., $p=5$). In fact, it also happens that there are counterexamples for fourth-powers, though this would only be found two decades later: Elkies developed a construction of infinitely many counterexamples, and did a computer search to find a concrete one:

Later, the smallest possible counterexample of a sum of three fourth-powers giving a fourth-power was discovered:

Of course, for fourth-powers, having three terms on the left-hand-side (LHS) is best possible because two is impossible by FLT. By contrast, for fifth-powers, we don't know whether four terms on the LHS is optimal: two is impossible, but three is open.

Also note that, while we know infinitely many counterexamples for fourth-powers, we only know two counterexamples for fifth-powers (including the one given above) if we restrict to positive integers and if we disallow solutions where all terms have a common factor. If we allow negative integers, we know precisely two more counterexamples; I'll come back to this issue of negative numbers later.

What about sixth-powers? We don't know. More accurately, we don't know whether it is possible to use strictly less than 6 terms on the left-hand side (thus giving other counterexamples to Euler's conjecture): in fact we don't know of counterexamples to Euler's conjecture with other powers than 4 or 5 (and for 5 as I explained we only know a handful). But there's more: we also don't know whether we can write a sixth-power as a sum of six sixth-powers; the best known expression is a sum of seven sixth-powers. By contrast, we do know how to write a seventh-power as a sum of seven seventh-powers, and likewise for eight. For all larger powers, we no longer know; all known solutions use more terms than the value of the power.

To sum up the number of terms needed in the LHS:

• $p=2$: two terms possible
• $p=3$: two terms impossible by FLT, three terms possible
• $p=4$: two terms impossible by FLT, three terms possible
• $p=5$: two terms impossible by FLT, four terms possible, three terms open
• $p=6$: two terms impossible by FLT, seven terms possible, everything else open
• $p=7$: two terms impossible by FLT, seven terms possible, everything else open
• $p=8$: two terms impossible by FLT, eight terms possible, everything else open
• $p\ge 9$: two terms impossible by FLT, all known solutions involve strictly more than $p$ terms

Such results are reviewed here and more exhaustively by the venerable EulerNet project (table "KNOWN LOWER BOUNDS", column 1), a volunteer computing project using BOINC to search for solutions to these problems.

In terms of general results, as far as I can tell, there are no known lower bounds on how many terms are necessary beyond what is given by FLT (i.e., two terms never suffice for $p>2$); and there is no known general upper bound on how many terms suffice as a function of $p$ except for what follows from Waring's problem (as I'll summarize below). See also the discussion here.

Equations $x_1^p+\cdots +x_k^p=y_1^p+\cdots +y_l^p$

A relaxation to the previous equation is to study equal sums of like powers, where both sides of the equation are sums of potentially more than one power (but all powers are still the same). Of course, we then look for nontrivial solutions, i.e., solutions where we cannot use the same number on both sides, and we want to find the smallest possible numbers $k$ and $l$ of terms on each side to have a solution. How large do these get, depending on the power $p$? Do we get solutions with a smaller total number of terms than in the case of the previous section (where we impose $l=1$)?

This rephrased problem is also more robust in presence of negative numbers. Allowing negative numbers of course has no impact when $p$ is even, but when $p$ is odd it could make a difference. However, when $k$ and $l$ can be greater than 1, then we can always make all numbers positive up to moving them to the other side of the equation. For this reason, allowing negative numbers makes no difference in the current section, whereas allowing them in the previous section reduces (for odd $p$) to the current section.

So what do we know? There is the Lander, Parkin, and Selfridge conjecture which asserts that we must have at least $p$ total terms for power $p$, i.e., $p\le k+l$. Contrast this to Euler's refuted conjecture: the latter only talked about equations with one term in the RHS ($l=1$), but it claimed that solutions needed $p$ terms in the LHS, i.e., the stronger bound $p+1\le k+l$.

The Lander, Parkin and Selfridge (LPS) conjecture has not been refuted yet. None of the counterexamples of the previous section refute it, because they beat Euler's bound by just one term, i.e., they ensure $k+l=p$. By contrast, achieving a fifth-power as a sum of three fifth-powers would refute the LPS conjecture, because the total number of terms would be 4. We also do not know whether for $p=5$ we could have $l=k=2$, i.e., a nontrivial solution to $a^5+b^5=c^5+d^5$. This amounts to asking whether there is an integer that can be written as a sum of (positive) fifth-powers in two different ways, which is a long standing open problem that I'll mention again at the very end of the post.

The LPS conjecture is not interesting for squares, and for cubes it is true (and not tight) because of FLT: you need at least two terms on each side so at least 4 terms in total, whereas the LPS conjecture makes the weaker assertion that you need at least 3 terms. For fourth-powers and fifth-powers we know solutions matching exactly the bound (e.g., the counterexamples to Euler's conjecture from the previous section), and for sixth-powers we had no Euler counterexample in the previous section but we do know a solution with 3 terms of each side (i.e., with $k=l=3$):

See this Wikipedia list for other similar solutions.

More results are in the EulerNet table "KNOWN LOWER BOUNDS". In particular for sixth-powers we don't know a solution with two terms on one side (i.e., $k=2$ and $l=4$ or vice-versa). We also know according to this table that eight total terms are possible for eighth-powers, but it is open whether seven total terms suffice for seventh-powers -- and the higher the powers, the less we know.

As far as I can see, we have no general lower bounds whatsoever on this problem, i.e., we have no weakening of the conjecture which is known to be true, except again from FLT when requiring one side to have only one term ($l=1$). In the current state of affairs no proven results seem to rule out that there could exist solutions to $a^5+b^5=c^5+d^5$, and more generally to $a^p+b^p=c^p+d^p$ for every $p\ge 5$. It also seems that we don't know whether there is any difference in the number of terms required when imposing $l=1$ like in the previous section, compared to the general case -- though as I explained we know some expressions using $l>1$ for which there is currently no known analogue with the same $k+l$ but with $l=1$.

Also, we do have bounds on the related Prouhet-Tarry-Escott problem, but I'm not covering this one here because it involves equations with multiple powers at the same time (see also this for multiple powers).

As for upper bounds, as far as I understand the only ones that we do have are the ones deriving from Waring's problem, which is the next question I'll talk about.

Achieving all numbers as $x_1^p+\cdots +x_k^p$

So: instead of finding sums of a small number of $p$-th powers that are equal to a $p$-th power, we could instead wonder how many terms are needed to express all positive numbers as a sum of that number of $p$-th powers. This is related to equations of the form $x_1^p+\cdots +x_k^p=y^p$ from the first section: if the LHS ranges over all numbers, then in particular it ranges over $p$-th powers, so the equation with $y^p$ on the RHS will necessarily have a solution.

For squares ($p=2$), the situation seems reasonably well-understood. We understand:

• Which numbers can be written as the sum of two squares;
• Which numbers can be written as the sum of three squares;
• That all numbers can be written as the sum of four squares.

For this last point, we allow the squares to be zero. If we only allow nonzero squares, we still get all numbers but with a small infinite list of exceptions¹, and allowing five nonzero squares suffices to achieve every number greater than 33. See this paper for a more general discussion of what happens when we disallow zero terms; I won't discuss the issue further and will always assume that zero is allowed.

For cubes ($p=3$), the situation immediately gets more messy:

• For numbers that can be written as the sum of two cubes, there is a Wikipedia page and a paper giving a characterization
• We do not know which numbers can be written as the sum of three cubes
• What we do know is that every natural number is a sum of at most 9 cubes of positive numbers, and 9 terms is the smallest possible number of terms in general: see Waring's problem

What about higher powers? Well, Waring's problem precisely asks us to figure out, for each power $p$, the minimal number of terms $g\left(p\right)$ that we may need in sums of (possibly zero) $p$-th powers in order to achieve all integers. We know that the function $g$ is well-defined in the sense that for every $p$ there is a finite number of terms $g\left(p\right)$ that suffices², and there is even an explicit formula that is conjectured to give the value of $g\left(p\right)$ for arbitrary $p$.

There is also a variation where, instead, we ask for the number $G\left(p\right)$ of terms in sums of $p$-th powers that allow us to achieve all sufficiently large integers. This is far less understood: we do know that this function is well-defined and admits upper bounds, but the exact values are unknown except for squares (where we know from the above characterizations that $G\left(2\right)=g\left(2\right)=4$) and for fourth-powers (where $G\left(4\right)=16$ is known). For cubes we do not know how many terms are needed: we could achieve all numbers with $g\left(3\right)=9$ terms, but we know that 7 cubes suffices for all but finitely many numbers; however this does not match the best-known lower bound of $4$. So $4\le G\left(3\right)\le 7$ and the exact value is not known.

There is another variation: we can allow negative numbers, and more generally (to have a version which is also interesting for even powers) we can ask how many terms we need to achieve all integers when we also allow minus signs in front of the terms. This is the so-called easier Waring's problem. It is "easier" in the sense that it is less difficult to prove that indeed for every power there is a number of terms that will always suffice, but it is otherwise less understood than the original problem described earlier. For instance, when allowing minus signs, we do not know whether 4 terms are needed for $p=3$, i.e., whether every integer can be written as a sum of four cubes of (possibly negative) integers; as this paper reviews, we know that 3 terms are not enough and that 5 terms suffice.

It then seems natural to combine both variants and ask for the least number of terms (positive and negative) that is needed to achieve all numbers except a finite number of exceptions. This variant appears under the name $W(\cdots )$ on page 6 of these MIT course notes, but it does not appear to have been otherwise studied.

How many times can we achieve numbers as a sum of like powers?

Beyond the study of how many terms are required in sums of like powers to cover all integers (or almost all of them), there is the question of the number of different ways in which we can express every integer.

For squares, we have some understanding of the number of possible ways in which numbers can be written as sums of squares, and of the number of ways in which they can be written as sums of 3 squares (with zero being allowed).

For cubes, again, our understanding is more sparse. The question relates to the famous anecdote of the taxicab number between Hardy and Ramanujan about the fact that 1729 is the smallest number that can be expressed as the sum of two positive cubes in two different ways. (Note that if we allow negative numbers then 1729 is no longer the smallest, because 91 can be expressed as $3^3+4^3$ and as $6^3-5^3$, see here.) The sequence of taxicab numbers studies more generally the smallest numbers $Ta\left(n\right)$ that can be written as sums of cubes in $n$ different ways.

The question extends to generalized taxicab numbers, with $Taxicab(p,j,n)$ being the smallest integer that can be written as the sum of $j$ $p$-th powers in $n$ different ways. Of course we know very little about these, and they happen to be related to the conjecture that we saw earlier by Lander, Parkin and Selfridge. Indeed, nontrivial solutions to $x_1^p+\cdots +x_k^p=y_1^p+\cdots +y_l^p$ with $k=l$ are in fact giving us a way to write the same number as sums of $k$ $p$-th powers in two different ways. Thus, the LPS conjecture would imply that $Taxicab(p,j,2)$ is undefined unless $2j\ge p$. In particular one open question is whether $Taxicab(5,2,2)$ is defined at all: this is again the open question of whether there is an integer that can be written as a sum of two positive fifth-powers in two different ways.