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# Prove that 49 is the only prime square to be followed by twice a prime square and then a semiprime

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Let $\tau(n)$ denote the number of divisors of $n$. OEIS sequence A309981 gives the smallest $k$ such that the tuple $(\tau(n), \tau(n+1), \ldots, \tau(n+k))$ uniquely determines $n$.

For small $n$ the value can often be verified by case analysis in residues to a suitable modulus, but $n=49$ is more resistent (and the notes in the OEIS history show that the person who posted the sequence considered it more challenging). No reference is given, and the correctness of $a(49) = 2$ is in the examples essentially as a bare assertion.

How can it be shown that $\tau(n) = 3$, $\tau(n+1) = 6$ and $\tau(n+2) = 4$ has a unique solution?

For comparison, and in case it's helpful, I give the standard case analysis.

$n$ must be a prime square $p_0^2$; it's not $2^2$ because $\tau(2^2 + 1) = 2 \neq 6$, so $n$ is odd. It's not $3^2$ because $\tau(3^2 + 1) = 4 \neq 6$. Therefore $p_0$ is coprime to $6$ and $n \equiv 1 \pmod 6$. Note also that since it's an odd square, $n \equiv 1 \pmod 8$. Combining the two, $n \equiv 1 \pmod {24}$.

$n+1$ is either $p_1^5$ or $p_1^2 q_1$. It's even; it's not $32$ because $\tau(31) = 2 \neq 3$, so it's either $4 q_1$ or $2 p_1^2$. But $4 q_1 \not \equiv 2 \pmod {24}$ so $n+1 = 2p_1^2$.

$n+2$ is either a prime cube $p_2^3$ or a semiprime $p_2 q_2$. Since $n+2 \equiv 3 \pmod {24}$ it's divisible by $3$; if it's a prime cube then it's $3^3$, but $\tau(3^3 - 1) = 4 \neq 6$, so $n+2 = 3 p_2$.

In summary \begin{array}{ccccc}n &=& p_0^2 &\equiv& 1 \pmod {24} \\ n + 1 &=& 2p_1^2 \\ n + 2 &=& 3p_3 \end{array}

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No other solutions $n\leq10^{200}$. (1 comment)
Source for the sparsity of candidates (1 comment)

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Not an answer, but too long for a comment:

Suppose $n$ is a positive integer such that $\tau(n)=3$ and $\tau(n+1)=6$ and $\tau(n+2)=4$. You already note that then $n=p^2$ for some prime number $p$, and $n+1=2q^2$ for some prime number $q$. It follows that $$p^2-2q^2=-1,$$ which has the form of a Pell equation. It follows that there is an integer $k$ such that $$p+q\sqrt{2}=(1+\sqrt{2})(3+2\sqrt{2})^k=(1+\sqrt{2})^{2k+1}.$$ Explicitly the coefficients of the right hand side are given by

\begin{array}{ccc} x_k &=& \frac{(1+\sqrt{2})^{1-2k}-(1+\sqrt{2})^{2k-1}}{2},\\ y_k &=& \frac{(1+\sqrt{2})^{1-2k}+(1+\sqrt{2})^{2k-1}}{2\sqrt{2}}. \end{array}

In particular we see that if $2k-1$ is not prime, say $2k-1=ab$ with $a,b>1$, then $x_k$ has two factors $$(1+\sqrt{2})^a-(1+\sqrt{2})^a \qquad\text{ and }\qquad (1+\sqrt{2})^b-(1+\sqrt{2})^b,$$ which are both greater than $2$, and so $x_k$ is not prime, and similarly $y_k$ is not prime. This means there exists a prime number $r$ such that

\begin{array}{ccc} p &=& \frac{(1+\sqrt{2})^r-(1+\sqrt{2})^{-r}}{2},\\ q &=& \frac{(1+\sqrt{2})^r+(1+\sqrt{2})^{-r}}{2\sqrt{2}}. \end{array}

A brute force check shows that there is no other solution $n$ with $n\leq10^{200}$.

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