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Are there infinitely many squares which cannot be written as the sum of a smaller non-zero square and twice a triangular number?

In other words, is the list given at https://oeis.org/A230312 infinite? And if yes, can you explain/cite its proof please?

Thanks in advance!

Ok-Virus2237
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1 Answers1

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The triangular number $T(b)$ is $(b^2+b)/2$, so a number is twice a triangular number iff it has the form $b^2+b$.

The OEIS entry you link to provides the answer, in the form of a conjecture credited to Alonso del Arte in 2013: that a number $n^2$ is in A230312 iff $4n^2+1$ is prime.

Let $a, b, n$ be non-negative integers. Say "$(a, b)$ solves $n^2$" to mean $n^2=a^2+2T(b)$. Clearly $(n, 0)$ solves $n^2$. A square $n^2$ has your criterion iff $n^2$ cannot be solved in any other way. $(a, b)$ solves $n^2$ iff

\begin{align*} n^2 &= a^2 + b^2 + b\\ \iff 4n^2+1 &= 4a^2 + 4b^2 + 4b + 1\\ &= (2a)^2 + (2b+1)^2 \end{align*}

If $4n^2+1$ is prime, it is the sum of two squares, but in only one way. Thus only one ordered pair $(a, b)$ solves $n^2$. This must be $(n, 0)$. Therefore $n^2$ has your criterion. On the other hand, if $4n^2+1$ is composite, it is the sum of two squares in at least two ways, and some $(a, b)$ with $a<n$ and $b>0$ solves $n^2$, so $n^2$ does not have your criterion. This, I believe, proves del Arte's conjecture.

So the question you ask amounts to "are there infinitely many integers $n$ where $4n^2+1$ is prime?". As Eric Snyder said, We probably don't know..

Rosie F
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  • So is it like if we manage to prove that the list given at https://oeis.org/A230312 is infinite, in other words, there are infinitely many squares which cannot be written as the sum of a smaller non-zero square and twice a triangular number, then as a consequence we will prove that there are infinitely many primes of the form $k^2+1$ which is basically Landau's 4th problem? Am I right? – Ok-Virus2237 Mar 13 '24 at 19:09