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Below is my original question, which has since been modified to a more general form.

Prove that $\forall p,q \in \Bbb P$ and $k \in \Bbb Z^+$ such that $q \equiv -1 \bmod 3$ and $p \equiv 1 \bmod 3, 2^{2k-1}q$ and $2^{2k}p$ are non-trivial polygonal numbers, where the non-trivial polygonal numbers are defined

$P_s(n)=s \dfrac{(n+1)(n+2)}2+n+2$, with $n,s \in \Bbb Z^+$.

Also prove, for the remaining case, that $2^k3$ is always a non-trivial polygonal number.

This conjecture is related to my open question here.

Update:

Further tests have suggested that we can make the following more general statements

$\forall m,j,k \in \Bbb Z^+$, $3^j(m+2)^k$ and $(m+2)^{k+1}$ are non-trivial polygonal numbers,

$\forall a,b,c,d,j,k \in \Bbb Z^+$ such that $a=2c+1$ and $b=d\dfrac{a-1}2+1=dc+1$, $a^jb^k$ is a non-trivial polygonal number. Note that this gives as a special case $2^k3^j$.

Generalizing the original statement,

$\forall a,b,k \in \Bbb Z^+$ such that $a \equiv -1 \bmod 3$ and $b \equiv 1 \bmod 3$, $2^{2k-1}a$ and $2^{2k}b$ are non-trivial polygonal numbers, with the exception $a=2,k=1$, which is also an exception for the original statement.

This may be too much to ask for in a single answer, and if so we can talk about what to do about the others once a proof has been produced of at least one of the statements. I've checked all of them using the polygonal numbers up to $760000$.

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Jaycob Coleman
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1 Answers1

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The problem turns out to be much simpler than it looked at the first glance. Most of the cases discussed are straightforward due to the fact that polygonal numbers are quite common among the integers (especially for small values of $n$).

Lemma: Let $r,t\in\mathbb{Z}^+$ be positive integers such that ${r\equiv 1}\pmod{t}$. Then, the integer $m=(2t+1)r$ is a polygonal number (non-trivial if $r>1$).
Proof: Set $s=\frac{r-1}{t}=\frac{1}{t}\left(\frac{m}{2t+1}-1\right)$ and calculate the number $P_s(2t-1)=m$.

Applying it to...

  • ... $t=1$ tells us that any multiple of $3$ is a polygonal number (and they're non-trivial beginning with 6). This resolves the question of numbers of the form $2^k3$ and also $3^j(m+2)^k$.
  • ... $t=c$ answers the question of $a^jb^k$ (with $a=2c+1$ and $b=dc+1$) in affirmative too.

Lemma: Let ${t\equiv 2}\pmod{3}$ be a positive integer. Then, $m=2t$ is a polygonal number (non-trivial for $t>2$).
Proof: Set $s=\frac{t-2}{3}$ and get $P_s(2)=m$.

Applications:

  • Set $t=2^{2k-2}a$ for ${a\equiv -1}\pmod{3}$ to see that $2^{2k-1}a$ is a polygonal number (non-trivial if $k>1$ or $a>2$).
  • Set $t=2^{2k-1}b$ for ${b\equiv 1}\pmod{3}$ to get $2^{2k}b$ being polygonal number (non-trivial for $k>1$ or $b>1$).

Lemma: If $b,e\in\mathbb{Z}^+$ are positive integers with $b>2$, then $m=b^e$ is a polygonal number (non-trivial if $e>1$).
Proof: Set $s=2\frac{b^{e-1}-1}{b-1}$ (this is an integer) and obtain $P_s(b-2)=m$.

This resolves the question of $(m+2)^{k+1}$.

Peter Košinár
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