I am very interested by problem in mathematics which are difficult (go at least 10 years without a resolution, say) but which have a solution that is short and elementary.
In this video Launay gave an example, namely A non-terminating game of Beggar-My-Neighbor (incorrectly claimed in Winning Ways to have been solved a long time ago).
Launay calls these problems "Pokémon problems": difficult to catch but once you catch them, they're in your pocket.
Question: What are other problems like this?
The question, raised by Euler, as to whether fewer than $n$ $n$th powers could sum to an $n$th power, was settled considerably more than ten years later by Lander and Parkin, the justification taking only one line, at high school level, $$ 27^5+84^5+110^5+133^5=144^5 $$ see
For reasons I'll explain at the end, I'm not certain this one qualifies, but here goes:
In 1934, Romanoff (sometimes spelled "Romanov") proved that the integers of the form $2^k+p$ ($p$ being a prime) have positive density. He asked whether there are infinitely many odd numbers that are not of the form $2^k+p$. In 1950, Erdős was the first to give the affirmative answer to this question; in fact, he produced an arithmetic progression of odd numbers, none of them of the form $2^k+p$.
His proof, in its entirety, is this:
"Clearly every integer satisfies at least one of the following congruences: $0\bmod2$, $0\bmod3$, $1\bmod4$, $3\bmod8$, $7\bmod{12}$, $23\bmod{24}$. Therefore, if $x$ is congruent to $1\bmod2$, $1\bmod7$, $2\bmod5$, $2^3\bmod{17}$, $2^7\bmod{13}$, and $2^{23}\bmod{241}$, then for any $k$, $x-2^k$ is a multiple of (at least) one of the primes $3$, $5$, $7$, $13$, $17$, $241$."
To see how this works, say, for example, that $k\equiv3\bmod8$. Then $2^k\equiv2^3\equiv x\bmod{17}$, so $x-2^k$ is a multiple of $17$, so there is no prime $p$ such that $x=2^k+p$.
By the Chinese Remainder Theorem, there is an infinite arithmetic progression of odd numbers $x$ satisfying all six of the given congruences, hence, not of the form $2^k+p$, $p$ prime.
Odd numbers not of form $p + 2^k$ are tabulated at https://oeis.org/A006285 along with many links to the literature.
The Erdős paper is On integers of the form $2^k+p$ and some related problems, Summa Brasiliensis Mathematicae 2 (1950) 113-123, available at https://users.renyi.hu/~p_erdos/1950-07.pdf.
Now, here's why I'm not certain that this qualifies. I don't know when Romanoff asked whether there were infinitely many odd numbers not of the form $2^k+p$. I haven't actually read the 1934 paper, so I don't know whether the question is there. Erdős doesn't cite the 1934 paper for the question; he just writes, "personal communication". So, it's possible that fewer than ten years passed between the publication of the question, and the solution by Erdős.
The question, raised by Euler, of whether there exists a pair of orthogonal latin squares of order $n$ for some $n\equiv2\bmod4$, was settled (in the affirmative) considerably more than ten years later, by Bose and Shrikhande, in the case $n=22$. Parker then found an example for $n=10$, which I guess counts as a ten-line proof. A picture is at https://upload.wikimedia.org/wikipedia/commons/d/d7/Scientific_American_November_1959_Graeco_Latin_square.svg
Lomonosov's theorem that a bounded operator on a complex (infinite-dimensional) Banach space that commutes with a nontrivial compact operator has a nontrivial invariant subspace had a surprisingly simple proof.
Incidentally, a very similar question was asked on hsm.stackexchange.com some time ago: Is there any example of a long-standing mathematical conjecture whose resolution did not require advanced knowledge? There are many good answers there which I will not repeat here.
Problems like "Is this number (which is not congruent to 4 or 5 mod 9) the sum of 3 cubes of integers?" have a short answer (if the answer is yes) but it is very hard to find the answer.
\begin{align} \textbf{Q: } & \text{ What is the prime factorization of } 2^{67}-1\text{?} \\[4pt] \textbf{A: } & ~193{,}707{,}721 \times 761{,}838{,}257{,}287. \end{align} Everybody knows that Frank Nelson Cole did this by hand. (Except those who don't.)
