$P = b^\frac{8}{11}a^\frac{20}{11} \in \mathbb{N}$, with $a, b \in \mathbb{N}, a, b > 1$. Find smallest such $P.$

Question

I was solving a problem and got to a point where I had to find the smallest $P = b^\frac{8}{11}a^\frac{20}{11} \in \mathbb{N}$, with $a, b \in \mathbb{N}$ and $a, b > 1.$ I tried different values using a calculator and found that $(a, b) = (2, 8)$ seems to produce the smallest $P (= 16).$ However, I struggled to argue why this is the case.

$P = \sqrt[11]{b^8a^{20}}.$ Since $a > 1$ and $a$ assumes the higher exponent, I was tempted to think that $a$ must be 2 for $P$ to be the smallest. Then $P = 2\sqrt[11]{b^82^{9}} \implies b^82^9 = c^{11} \implies c = 2n \implies b = \sqrt[8] {2^2n}$ and $n$ is even. With that, I could find that $n = 4$ is the smallest value that gives $b \in \mathbb{N}$ ($b = 8$).

However, I'm not confident in this argument, because there might exist a case where $a > 2$ and $b < 8$ such that $P < 16,$ but I'm not sure how to argue that this doesn't exist. Please advise.

Thank you!

EDIT: After some thought, I realize it could be argued that suppose there exist $a > 2$ and $b< 8$ such that $P < 16.$ Then $b^8a^{20} = P^{11} < 16^{11} \implies a^{20} < \frac{16^{11}}{b^8} \leq \frac{16^{11}}{2^8} \implies a < 3.48 \implies a = 3,$ in which case $b \notin \mathbb{N}.$ Thus, the original assumption is false. However, I still wondered if there exists a shorter argument overall for $P = 16.$

Answer 1

We first claim that a feasible pair $(a,b)$ must be a power of prime. Assume it is not, then we write \begin{align*}a&=p_1^{k_1}\cdots p_m^{k_m}\\ b&=q_1^{j_1}\cdots q_n^{j_n}\end{align*} as $a,b>1$. If there exists $i_w,j_w\equiv 0\mod11$, then we can remove them as we want the smallest pairs, so we can assume none of them is $\equiv 0$ in modulo $11$. Then this means $a,b$ must have every prime factor the same. Hence, we may write $q_w=p_w$ for simplicity. Now since $P\in\mathbb{N}$, we have for each $w$, $$20k_w+8j_w\equiv 0\mod 11\tag{*}$$ But since we want smallest, we can drop away all other prime factors as $(p_1^{k_1},p_1^{j_1})$ is already a feasible choice. Hence, we must have $a,b$ are power of prime. This means the problem of minimising $P$ reduced to minimising $k_1+j_1$.

As a corollary, this confirms that your assumption is true, in the sense we should pick $p_1=2$. Otherwise it must be larger than $p_1^{k_1+j_1}$ (Note that $k_i,j_i$ are independent from your choice of $p_i$, they just need to satisfy $(*)$). Hence, we have $(a,b)=(2^{k},2^j)$ subject to $8k+20j\equiv 0\mod 11,k,j\in\mathbb{N}$. Clearly $8k+20j\ge28$. We use trial and error to test the smallest solution.

• $8k+20j=33$: This is nonsense as LHS is even but RHS is odd.
• $8k+20j=44$: This can be reduced to $2k+5j=11$. Either by Euclidean Algorithm, or by observation, we have $2(3)+5(1)=11$. So we are done. In sum, $(a,b)=(2^3,2^1)=(8,2)$ gives the smallest solution.