Levels 14 & 28

For certain even levels divisible by $7$, the situation is now different. We can still find an eta quotient $a$ such that $a+m = b$ is also an eta quotient, but only for one rational $m$.

I. Moonshine functions: Define $$a(\tau) =\left(\frac{d_1\,d_7}{d_4\,d_{28}}\right),\quad b(\tau) = \left(\frac{d_2^3\,d_{14}^3}{d_1\,d_4^2\,d_7\,d_{28}^2}\right)$$ then $a+2 = b$ or, $$\left(\frac{d_1\,d_7}{d_4\,d_{28}}\right)+2 = \left(\frac{d_2^3\,d_{14}^3}{d_1\,d_4^2\,d_7\,d_{28}^2}\right)$$  for the unique $m=2$. This is the single trinomial identity of level 28. First, we have the 3 moonshine functions of level 14,
$$\begin{align} j_{14A}(\tau) &= \left(\sqrt{j_{14C}}+\frac{1}{\sqrt{j_{14C}}}\right)^2\\ j_{14B}(\tau) &= \left(\frac{d_1\,d_7}{d_2\,d_{14}}\right)^3\\ &= b+\frac{4}{b}-4\\ j_{14C}(\tau) &= \left(\frac{d_2\,d_7}{d_1\,d_{14}}\right)^4 \end{align}$$ and the 4 moonshine functions of level 28,
$$\begin{align}\quad j_{28A}(\tau) &= \sqrt{j_{14A}(2\tau)} = j_{28D} (\tau)+\frac{1}{j_{28D} (\tau)}\\ j_{28B}(\tau) &= \left(\frac{d_2^2\,d_{14}^2}{d_1\,d_4\,d_7\,d_{28}}\right)^3\\ &= a+\frac{4}{a}+4\\ \color{red}{j_{28C}}(\tau) &=  \frac{d_1\,d_7}{d_4\,d_{28}}\;=\; a\\ j_{28D}(\tau) &=  \sqrt{j_{14C}(2\tau)} = \left(\frac{d_4\,d_{14}}{d_2\,d_{28}}\right)^2 \end{align}$$ Note that $j_{14B}\big(\tfrac12+\tau\big) = -j_{28B}(\tau)$.