Let m be an arbitrary whole number, and d be a whole number value that we wish to deduct sequentially from m, until we find the first example of a result that is non-trivially polygonal as a plane figurate number.
If the test for polygonality examines and then dismisses certain possibilities of regular polygon, can this information be used to categorically exclude certain others without us having to test for them explicitly?
Given that certain values may be encountered that will not succeed as being non-trivially polygonal, would the exercise only be bounded by m(mod d) and a test of the residue?
Finally, how does this compare as an exercise with say, dividing by primes up to the square root of a whole number?
Edit
In response to the comments received, the question is meant to focus on identifying the non-centred varieties of regular convex polygon of order greater than 2. Whether an arbitrary number is capable of being polygonal in this sense can be determined by application of an elementary "yes/no" square conversion test. This involves a root-finding regime which may be considered expensive, but it is an elementary exercise. What we are asking about, is how repeated application of this routine can be aborted or potentially sped up if we are iterating through candidate possibilities to achieve a first "hit". Moreover, does it become computationally infeasible to pursue repeated deductions from m when d and m are at some particular "distance" from each other to start with?
Edit 10/10/21
The motivation behind asking the main question "How practical is it to test whether a particular number is polygonal?", is that it appears to be an open problem as to how to do this efficiently. We can employ individual tests against any particular arbitrary number to discover if it is n-gonal in a non-trivial way. This is time consuming. If we were able to identify more readily, the potential polygonality characteristics "locked up" as it were, in a certain number, it would essentially provide an instantaneous route to being able to factor a value for which our m above, stands as an index point. It would be helpful to be able to employ a device such as one that says "if this number is not polygonal in this way, it will not (or most probably will not) be polygonal in such and such another way".
It is striking that questions concerning polygonal numbers in general, are somewhat "rare", both here on MSE and Math Overflow. Indeed, there are contributions from well regarded individuals that seem to support the view that beyond triangular, square and pentagonal numbers there is no reason to expect that other plane figurate numbers have much utility beyond being visual peculiarities. It might be a surprise to discover therefore that our inability to wrestle with them expeditiously detracts from an otherwise valuable route in dispelling the notion that factoring is necessarily a hard problem.
Any help, comments or advice received in relation to this question will serve a useful purpose, even if the state of current knowledge and opinion is that it continues to be impractical to identify the polygonality traits of a certain number without exercising undue labour.
Edit 11/10/21
When evaluating the processes involved that determine the degree to which it is practical to repeatedly test a value for polygonality, it might be instructive to consider an illustration of what is alluded to in the notion of "distance" above. A nice example would be:
m = 340549
d = 367
Eventually, repeated subtraction would reveal the value 280728.
This is a 328-gonal number.
It will allow us to factor the semi-prime 115 973 281 219 without using division, which is to say that the factors
2089 and 55516171
are instantly available by elementary algebra.
Let us attempt to find a polygonal representation for $14$ using the formulas in Wikipedia.
The formulas may be summarized by the expression
$2N =2r+kr(r-1)$
where $k+2$ is the number of sides in the polygon and $r$ is the order of the polygonal representation. Order $2$ means the polygon has as many sides as the value of the number, which is the trivial representation.
For $14$ this gives
$2r+kr(r-1)=28$
$kr^2+(2-k)r-28=0$
Setting $k$ to a positive integer we require this equation to have rational roots for $r$, and so the discriminant
$(2-k)^2+112k=k^2+108k+4$
is to be a perfect square.
Complete te square in the quadratic expression $k^2+108k+4$ to get
$(k+54)^2-2912=m^2$
and so $2912$ must be rendered as a difference between two squares. We apply various factorization of $2912=2^5×7×13$ to get candidate differences. We go in order from factors that are close together to those that are farther apart.
$56×52 \implies 54^2-2^2, k+2=54-4+2=2$, no good.
$91×32 \implies$ no solution, factors must have the same parity to get a difference of integer squares
$104×28 \implies 66^2-38^2, k+2=66-54+2=14$, number of sides matches the given number, so the positive root for $r$ is $2$ and this is the trivial solution.
Factors farther apart than those giving the trivial solution would give polygons with more sides than the value of the number, a contradiction; therefore $14$ has only the trivial representation.
For a general value of $N$ the required value for the difference of squares would br $16N(N-1)$. A nontrivial solution then requires that this product have an even factor strictly between $2N$ and $4(N-1)$ with the complementary factor also even. Here for $N=14$ there were no such factors of $2912$ between $28$ and $52$.