A Poulet-number $n$ is a weak Fermat-pseudoprime to base $2$ , in other words a composite number $n$ with the property $$2^{n-1}\equiv 1\mod n$$
The first $34$ poulet-numbers of the form $k^2+1$ (positive integer $k$) are , according to my calculations :
$$[46657, 898705, 2433601, 23736385, 67371265, 96589585, 331240001, 878529601, 1370776577, 4294967297, 8019202501, 80542440001, 287210246401, 351596817937, 422240040001, 682160364901, 1086430982401, 1911029760001, 2359971888401, 13603040532901, 17767236614401, 18677955240001, 26042752304101, 364005569944901, 458631349862401, 598865079758401, 45824890274900101, 189623338816064401, 286245437364810001, 1148717191415062501, 2333246290710627601, 6017402415698251777, 18446744073709551617, 52807456278501210001]$$
They are all squarefree.
Questions :
- Is my list of the first $34$ poulet-numbers of the form $k^2+1$ correct ?
- Is every poulet-number of the form $k^2+1$ squarefree ?
- Are there infinite many Poulet-numbers of the form $k^2+1$ ?
Trial :
Assume $p$ is a prime with $p^2\mid k^2+1$ and $2^{k^2}\equiv 1\mod (k^2+1)$. Then , we have $$2^{k^2}\equiv 1\mod p^2$$ The order $u$ of $2$ modulo $p^2$ is therefore a divisor of $k^2$ , but because of $p\nmid k^2$ we have $p\nmid u$. Therefore we have $u\mid p-1$ , hence $p$ is a Wieferich prime. Can we finish the proof ? If yes, how ?