A Suguru puzzle consists of a bunch of polynominoes fit together in a rectangular shape. The goal of the puzzle is to fill all the cells in all the polynominoes with the numbers $1$ to the amount of cells the polynomino has. A pentomino requires you to fill in $1$, $2$, $3$, $4$, and $5$, for example. A number is not allowed to touch the same number, not diagonally either!
Recently I was wondering if it is possible to construct a Suguru puzzle without any given clues (no given numbers) with exactly one solution. I was able to find some examples so I started wondering for which sizes this would work. So my question is:
For which $a$, $b$ is it possible to construct an $a\times b$ Suguru puzzle with no given clues and exactly one solution?
I was able to prove some cases, but not all cases.
I was able to prove that if $a$ and $b$ are both odd, than there exists a possible arrangement with no given clues and exactly one solution unless both $a$ and $b$ are equal to $3$, in which case it is impossible to construct one.
If one side is even and the other one is odd, I was able to proof that there exists an arrangement if the even side is at least $6$ and the odd side at least $3$. If the odd side is equal to $1$ than no arrangement with the other side being even exists.
If the even side is equal to either $2$ or $4$, I am not sure whether it will ever be possible to construct a puzzle with no given clues and exactly one solution. So far I have not been able to find any working examples, so I believe there aren't any at all.
With the help of the 8x8 configuration which was provided by Richard Tobin, I was also able to prove that an arrangement exists if $a$ and $b$ are both even and at least 8, however, I still haven't figured out what happens if $a$ and $b$ are both even, but not both at least 8.
This brings all the unsolved cases down to three categories. $2\times n$ for all $n$, $4\times n$ for all $n$ and $6\times n$ for all even $n$
Does anyone know how I could prove all cases, or just some of the cases I have not yet already solved?