On finite products of $\frac{p+4}{p+2}$ with $p$ prime

Question

Let us consider factorizations of rational numbers greater than one. For integers $a>b>0$, clearly $$\frac ab =\prod_{n=b}^{a-1}\frac{n+1}n.$$ In view of Question 476578 and Max Alekseyev's comment there, it seems that each rational number $r>1$ can be written as a product of distinct elements of the set $$S_1=\left\{\frac{n+1}n:\ n\in\mathbb Z^+\ \text{and}\ 2n+1\ \text{is prime}\right\}=\left\{\frac{p+1}{p-1}:\ p\ \text{is an odd prime}\right\},$$ where $\mathbb Z^+$ denotes the set of all positive integers.

For integers $m>k\ge0$, it is apparent that $$\frac{2m+1}{2k+1}=\prod_{n=k+1}^m\frac{2n+1}{2n-1}.$$ Motivated by the set $S_1$, we introduce the set $$S_2=\left\{\frac{2n+1}{2n-1}:\ n\in\mathbb Z^+\ \text{and}\ 2n-3\ \text{is prime}\right\}=\left\{\frac{p+4}{p+2}:\ p\ \text{is an odd prime}\right\}.$$

Question 1. Whether for any $n\in\mathbb Z^+$ we can write $(2n+1)/(2n-1)$ as a product of distinct elements of $S_2$?

Question 2. If Question 1 has a positive answer, whether for any integers $m>k\ge0$ we can write $(2m+1)/(2k+1)$ as a product of distinct elements of $S_2$?

For each $n\in\mathbb Z^+$, let $a(n)$ denote the least odd squarefree number $m>1$ such that $$\prod_{p\mid m}\frac{p+4}{p+2}=\frac{2n+1}{2n-1}.$$ I have found the exact values of $a(1),\ldots,a(20),a(22),\ldots,a(53)$. For example, $$a(1)=50234415=3\times5\times7\times11\times23\times31\times61,\ \ \ a(2)=1085=5\times7\times31,$$ $$a(48)= 165694433=131\times373\times3391\ \ \text{and}\ \ a(51)=424958987=113\times1129\times3331.$$ For the exact values of $a(3),\ldots,a(20)$, one may consult https://oeis.org/A375351. It seems that $a(n)$ is particularly large when $n$ is divisible by $3$. My computation indicates that $a(21)>1.2\times10^{10}$ and $a(54)>1.2\times10^9$. I conjecture that $a(n)$ exists for any $n\in\mathbb Z^+$, i.e., Question 1 should have an affirmative answer.

Question 3. What are the exact values of $a(21)$ and $a(54)$?

I think, this question will be answered soon by an expert good at computation.

Your comments are welcome!

Answer 1

$a(21) = 675790721971 = 113\times157\times271\times367\times383$. First, note by calculation that this is a valid value.

We will set $r = \omega(a(21))$.

Now from @Deyi Chen's answer we have $r \leq 5$. If $r = 5$, then we have $p_5 \leq 8467$. Now we can construct an instance of weighted 5-SUM: for any prime $p$, we have a weight of $\log(p)$, and a vector based on the factorization of $\frac{p+4}{p+2}$. We will give random weights to all primes to project the vectors to integers. While this may add extraneous solutions, that's not a problem for showing that there aren't any. Executing this C++ code shows that $675790721971$ is optimal for $p_r \le 10^4$, so in particular it's optimal for $r = 5$.

Now, we will consider $r < 5$.

Setting $f(p) = \frac{p+4}{p+2}$, note that it's a decreasing function, so for each $i$ we have $f(p_1) f(p_2) \cdots f(p_i) \leq \frac{43}{41} = f(p_1) f(p_2) \cdots f(p_r) \leq f(p_1) f(p_2) \cdots f(p_{i-1}) f(p_i)^{r-i+1}$. Additionally, note that if we have a guess for $p_1, p_2, \dots, p_{r-1}$ we can compute $f(p_r) = \frac{\frac{43}{41}}{f(p_1) \cdots f(p_{r-1})}$, and because $f(p)$ is reduced for prime $p$ we can just check if the numerator and denominator have difference 2, and if they do whether numerator-4 is prime.

The following Python code checks that and verifies that there are no such solutions:

from fractions import Fraction

from sympy import *

prims = list(primerange(40, 10**6))

f = lambda p: Fraction(prims[p] + 4, prims[p] + 2)

# r = 4
p1 = 0
while 43/41 <= f(p1)**4:
    assert f(p1) <= 43/41
    p2 = p1 + 1
    while 43/41 <= f(p1) * f(p2)**3:
        while 43/41 <= f(p1) * f(p2):
            p2 += 1
        p3 = p2 + 1
        while 43/41 <= f(p1) * f(p2) * f(p3)**2:
            while 43 / 41 <= f(p1) * f(p2) * f(p3):
                p3 += 1
            target = Fraction(43, 41) / (f(p1) * f(p2) * f(p3))
            if target.numerator - target.denominator == 2 and isprime(target.numerator - 4):
                print(p1, p2, p3, target)
                exit(0)
            p3 += 1
        p2 += 1
    p1 += 1

# r = 3
p1 = 0
while 43/41 <= f(p1)**3:
    assert f(p1) <= 43/41
    p2 = p1 + 1
    while 43/41 <= f(p1) * f(p2)**2:
        while 43/41 <= f(p1) * f(p2):
            p2 += 1
        target = Fraction(43, 41) / (f(p1) * f(p2))
        if target.numerator - target.denominator == 2 and isprime(target.numerator - 4):
            print(p1, p2, target)
            exit(0)
        p2 += 1
    p1 += 1

# r = 2
p1 = 0
while 43/41 <= f(p1)**2:
    assert f(p1) <= 43/41
    target = Fraction(43, 41) / f(p1)
    if target.numerator - target.denominator == 2 and isprime(target.numerator - 4):
        print(p1, target)
        exit(0)
    p1 += 1

# r = 1 is left as an exercise to the reader

```

Answer 2

$a(54)=1943156261=127\times653\times23431.$

Because $$ m=11854443675453271=41\times2089\times 4093\times4253\times7951$$ and $$\prod_{p\mid m}\frac{p+4}{p+2}=\frac{43}{41},$$ it implies that $$a(21)\leq11854443675453271<1.2\times10^{16}.$$ Claim 1. $\omega(a(21))\leq 7$ where $\omega(N)=\sum_{p\mid N}1.$
Proof. Let $\omega(a(21))=r$ and $a(21)=p_1p_2\cdots p_r,$ where $p_1<p_2<\cdots<p_r$. Noting that $$\frac{43}{41}=\prod_{i=1}^r\frac{p_i+4}{p_i+2}\geq\left(\frac{p_i+4}{p_i+2}\right)^i=\left(1+\frac{2}{p_i+2}\right)^i\geq 1+\frac{2i}{p_i+2},$$ we obtain $$p_i\geq41i-2$$ and $$1.2\times10^{16}>a(21)=\prod_{i=1}^{r}p_i\geq\prod_{i=1}^{r}(41i-2).$$ Since $\prod_{i=1}^{8}(41i-2)>2.8\times10^{17}$, it leads to $$r\leq 7.$$ Remark. It can be proven that if an odd squarefree number $m$ satisfies $\omega(m)=r$ and $\prod_{p\mid m}\frac{p+4}{p+2}=\frac{2n+1}{2n-1}$, then $m<(8(2n-1))^{2^r-1}.$

Claim 2. If $\omega(a(21))=5$ and $a(21)=p_1p_2\cdots p_5,$ where $p_1<p_2<\cdots<p_5$, then $p_5\leq 8467.$

Proof From Daniel Weber's comment, it follows that $$6.8\times10^{11}>a(21)=p_5\prod_{i=1}^{4}p_i\geq p_5\prod_{i=1}^{4}(41i-2)=61158240p_5.$$ So $$p_5<11119.$$ Since $$\frac{43}{41}=\prod_{i=1}^5\frac{p_i+4}{p_i+2}$$ there exist $1\leq i,j\leq5$ such that $$p_i+2 \equiv 0\pmod{41},$$ $$p_j+4 \equiv 0\pmod{43}.$$ If $i=j$ then $p_5\geq p_i=p_j\geq 14143$, this contradicts $p_5\leq 11119.$

Therefore $i\neq j$ and $p_i\geq367, p_j\geq 211.$ Hence, $p_4\geq 211$ and $$6.8\times10^{11}>a(21)=p_5p_4\prod_{i=1}^{3}p_i\geq 211p_5\prod_{i=1}^{3}(41i-2)=79656720p_5.$$ So $$p_5\leq 8467.$$

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a(75) exists. $$a(75)\leq507578951905423=449\times 557\times 727\times 1103\times 2531.$$

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Theorem 1. If an odd squarefree number $m$ satisfies $\omega(m)=r$ and $\prod_{p\mid m}\frac{p+4}{p+2}=\frac{2n+1}{2n-1}$, then $r \equiv 1 \pmod 2.$

Proof. Suppose $r \equiv 0 \pmod 2$, $m=\prod_i^rp_i$. In the set $\{p_1,\cdots, p_r\}$ there are $s$ numbers that are congruent to $1\pmod4$ and $ t$ numbers that are congruent to $-1\pmod4$.

Noting that$\frac{2n+1}{2n-1}=\prod_{i=1}^r\frac{p_i+4}{p_i+2}$, we have $$2n=\frac{\prod_{i=1}^r(p_i+4)+\prod_{i=1}^r(p_i+2)}{\prod_{i=1}^r(p_i+4)-\prod_{i=1}^r(p_i+2)}.$$ Clearly, $$\prod_{i=1}^r(p_i+4)+\prod_{i=1}^r(p_i+2)\equiv\prod_{i=1}^r(p_i+4)-\prod_{i=1}^r(p_i+2)\equiv0\pmod 2.$$ Since $s+t=r\equiv 0 \pmod 2$, we have $$\prod_{i=1}^r(p_i+4)+\prod_{i=1}^r(p_i+2)\equiv (-1)^t+3^s\equiv(-1)^t+(-1)^s\not\equiv0\pmod4 $$ and $$\prod_{i=1}^r(p_i+4)-\prod_{i=1}^r(p_i+2)\equiv (-1)^t-3^s\equiv(-1)^t-(-1)^s\equiv0\pmod4 .$$ Hence $$2n=\frac{\prod_{i=1}^r(p_i+4)+\prod_{i=1}^r(p_i+2)}{\prod_{i=1}^r(p_i+4)-\prod_{i=1}^r(p_i+2)} \not \in \mathbb{Z}.$$ This contradiction led to $r \equiv 1 \pmod 2.$ QED

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I have verified that $$a(75) = 507578951905423=449\times 557\times 727\times 1103\times 2531 \approx 5.08\times10^{14}.\tag{1}$$ We provide a proof for $(1)$. By Theorem 1, $\omega(a(75))$ is an odd number. Clearly, $\omega(a(75))\neq 1$. Similar to Claim 1, we have $\omega(a(75))\leq 5.$

Claim 3. $\omega(a(75))=5.$

Proof. We only need to prove that $\omega(a(75))\neq 3.$ Suppose $\omega(a(75))=3$ and $a(75)=p_1p_2p_3$ where $3\leq p_1<p_2<p_3$. Similar to Claim 1, we have $p_i\geq i(2n-1)-2=149i-2$. So $p_1\geq 149$ and $p_2\geq307.$

If $p_1\geq 443$, then $p_2\geq449$, $p_3\geq 457$ and $$\frac{151}{149}=\prod_{i=1}^3\frac{p_i+4}{p_i+2}\leq\frac{447}{445}\times\frac{453}{451}\times\frac{461}{459}=\frac{10372039}{10235445}<\frac{151}{149}.$$ Hence $$149\leq p_1\leq 439.$$

If $p_2\geq 22787$, then $p_3\geq22807$ and $$\frac{151}{149}=\prod_{i=1}^3\frac{p_i+4}{p_i+2}\leq\frac{153}{151}\times\frac{22791}{22789}\times\frac{22811}{22809}<\frac{151}{149}.$$ Hence $$307\leq p_2\leq 22783.$$

Therefore, under the conditions $149\leq p_1\leq 439$ and $307\leq p_2\leq 22783$, we solve for the prime solution $p_3$ of the equation $\frac{151}{149}=\prod_{i=1}^3\frac{p_i+4}{p_i+2}$. That is, check whether $$p_3=\frac{147 p_{1} p_{2}+890 p_{1}+890 p_{2}+4164}{p_{1} p_{2}-147 p_{1}-147 p_{2}-890}\tag{2}$$ is a prime number. After computer checks, the equation $(2)$ has no solution. Hence $\omega(a(75))\neq 3.$ QED

Claim 4. $$a(75) = 507578951905423=449\times 557\times 727\times 1103\times 2531.$$

Proof. By Claim 3, We have $\omega(a(75))=5$. Let $a(75)=\prod_{i=1}^5p_i$ where $3\leq p_1<p_2<\cdots<p_5.$ From $p_i\geq i(2n-1)-2=149i-2$, we obtain $$p_1\geq149,p_2\geq307,p_3\geq449,p_4\geq599.$$ We will now determine the upper bound for $p_i$ where $1\leq i \leq 4.$

If $p_1\geq 739$, then $p_2\geq743$, $p_3\geq 751$, $p_4\geq 757$, $p_5\geq 761$ and $$\frac{151}{149}=\prod_{i=1}^5\frac{p_i+4}{p_i+2}\leq\frac{743}{741}\times\frac{747}{745}\times\frac{755}{753}\times\frac{761}{759}\times\frac{765}{763}<\frac{151}{149}.$$ Hence $$149\leq p_1\leq 733.$$ Since $$507578951905423\geq a(75)=\prod_{i=1}^5p_i\geq149p_2(p_2+2)(p_2+4)(p_2+6),$$ $$507578951905423\geq a(75)=\prod_{i=1}^5p_i\geq149\times307p_3(p_3+2)(p_3+4),$$ $$507578951905423\geq a(75)=\prod_{i=1}^5p_i\geq149\times307\times449p_4(p_4+2),$$ we obtain $$p_2\leq1327,p_3\leq 2221,p_4\leq4969.$$

Therefore, under the conditions $149\leq p_1\leq 733$, $307\leq p_2\leq 1327$, $449\leq p_3\leq 2221$ and $599\leq p_4\leq 4969$, we solve for the prime solution $p_5$ of the equation $\frac{151}{149}=\prod_{i=1}^5\frac{p_i+4}{p_i+2}$. That is, check whether $$p_5= \frac{596\prod_{i=1}^4(p_i+4)-302\prod_{i=1}^4(p_i+2)}{151\prod_{i=1}^4(p_i+2)-149\prod_{i=1}^4(p_i+4)}\tag{3}$$ is a prime number. After computer verification and calculations, we obtain $$a(75) = 507578951905423=449\times 557\times 727\times 1103\times 2531.$$ QED