Erdős–Moser equation

The equality $1+2=3$ adds every positive integer below 3 to get 3. Squaring those same terms fails: $1^2+2^2=5$, while $3^2=9$.

Can any other endpoint and positive whole-number power succeed, when every smaller positive integer must be included? The Erdős–Moser conjecture says that $1+2=3$ is the only example. Selecting just a few smaller integers gives a different problem.[1]

Can raising every positive integer below an endpoint to the same positive integer power, then adding the results, give that power of the endpoint? The known success is $1+2=3$; Erdős–Moser predicts there is no other one. For squares at that endpoint, $1^2+2^2=5$, less than $3^2=9$.

With positive integers m and k, the full target is $$\sum_{j=1}^{m-1}j^k=m^k.$$ Every integer in that range contributes. The only proposed solution is $(m,k)=(3,1)$; the odd-exponent cases are known, leaving possible even-exponent solutions.

Here is one restriction with a visible mechanism. Moser's argument proves that any further solution has odd m and even k. Modulo 4, every odd integer to an even power is 1 and every even integer to that power is 0. For odd m, the sum has $(m-1)/2$ odd terms, whereas the right side is 1 modulo 4. Consequently $$\frac{m-1}{2}\equiv1\pmod4,\qquad m\equiv3\pmod8.$$ This is a necessary divisibility condition, not a completeness proof. The original equation still needs to be ruled out for every further possible endpoint and exponent.[1] [2]

The conjecture says no; even exponents remain unresolved.[1]

Strong lower bounds restrict any further even-exponent solution. August 2026 solutions to arithmetic-progression variants omit terms and therefore do not refute this conjecture.[1]

It tests how rigid an exact equality between a sum of whole-number powers and the next power can be. Complete lists of powers have systematic remainders when divided by primes, so an equality must obey many arithmetic restrictions at once. This brings addition of a whole list into contact with the multiplicative structure of the integers.[1] [2]

A selected pair of squares can match a third, as in $5^2+12^2=13^2$. The original question demands all twelve positive-integer squares below 13, so that familiar equality is not a solution. Including every term imposes a much more rigid arithmetic pattern than choosing a few summands.

Complete power sums have systematic remainders modulo primes. Matching one to a single next power must satisfy all those restrictions simultaneously, linking addition of many terms with prime divisibility. Moser's method uses such remainder identities to constrain any hypothetical solution. The interest is in how these local arithmetic constraints interact, beyond the observation that a finite search has not found another equality.[1] [2]