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Consider the following true statement. Given natural numbers n, m, s

(1) [tex]\sum_{i=0}^n a^i = \sum_{i=0}^s a^{(m+1)i} \sum_{i=0}^m a^i[/tex]

iff

(2) [tex]n+1=(m+1)(s+1)[/tex]

Now suppose that [tex]N[/tex] is a natural number where

(3) [tex]N = \sum_{i=0}^n a^i [/tex]

Now restrict [tex]a[/tex] to the natural numbers. Clearly for certain special numbers N we have a one to one map of factor pairs of [tex]N[/tex] and factor pairs of [tex]n+1[/tex] which is much smaller.

In order for this thereom to give non-trivial factorizations [tex]n+1[/tex] must not be prime. If this is true and (3) holds then [tex]N[/tex] can be factored easily.

If one thinks carefully about the algorithm for polynomial multiplication you can see that numbers defined by (3) above are quite special with respect to the base [tex]a[/tex].

In order to grasp the full implications of this theorem we need to think about two common algorithms.

1) The first is the base [tex]a[/tex] expansion algorithm used to write a natural number as a sum of powers of a. For example the number [tex]14[/tex] can be written as a sum of powers of [tex]3[/tex] as [tex]14 = 3^2+3+2[/tex]. In general this algorithm can be used to write

(4) [tex] N_a = c_k a^k + c_{k-1} a^{k-1} + ... + c_1 a + c_0[/tex]

Where some of the [tex]c_j[/tex] may be zero and all are less than [tex]a[/tex]. I will denote this algorithm by the capital letter [tex]\beta[/tex] so that [tex]\beta (N,a)[/tex] is equivalent to (4) above.

2) The second algorithm is polynomial multiplication. Now this algorithm proceeds by steps. We first multiply every term by every other term, then we collect terms. Denote this by [tex]P[/tex]. The difference between the result of this algorithm and the base [tex]a[/tex] expansion algorithm is best understood by looking closely at a third related algorithm.

3) The multiplication algorithm for natural numbers is really equivalent to polynomial multiplication taken one step further by carrying so that none of the digits are greater than the base and then evaluation of the resulting polynomial at the base value [tex]a[/tex]. Denote this carry step by [tex]C[/tex].

It follows that if [tex]N = ST[/tex] then [tex]\beta (N,a) = C(P(\beta (S,a),\beta (T,a)))[/tex] and under certain special conditions [tex]\beta(N,a)=P(\beta(S,a),\beta(T,a))[/tex]. One of these special conditions is the thereom stated above. This happens because of two things. First all coefficients of the summation terms on the right are equal to one in the base and the powers that do appear do not cause any duplication of powers, thus there is no carrying.

In the posts that follow I will try to highlight the implications of this. One of the first is that though it would appear that we are looking at a very restricted class of factorizations, in reality a shift operator can often be applied to [tex]\beta(N,a)[/tex] to arrive at a new base where coefficients are all one and the theorem applies. This is typically possible when [tex]\beta(N,a)[/tex] is not the result of carrying, or more formally there exists a pair of factors of [tex]N[/tex] such that the base [tex]a[/tex] expansion and subsequent multiplication produces a polynomial in [tex]a[/tex] with precisely the same coefficients as the base [tex]a[/tex] expansion of [tex]N[/tex].

Clearly this all plays around in an uncomfortable way with our notions of variables and constants. However if you look at it carefully you will see that there is merit here. Furthermore there exist shift operators that can transform certain base [tex]a[/tex] expansions into the form given in (1) above. I will talk about those next.

(1) [tex]\sum_{i=0}^n a^i = \sum_{i=0}^s a^{(m+1)i} \sum_{i=0}^m a^i[/tex]

iff

(2) [tex]n+1=(m+1)(s+1)[/tex]

Now suppose that [tex]N[/tex] is a natural number where

(3) [tex]N = \sum_{i=0}^n a^i [/tex]

Now restrict [tex]a[/tex] to the natural numbers. Clearly for certain special numbers N we have a one to one map of factor pairs of [tex]N[/tex] and factor pairs of [tex]n+1[/tex] which is much smaller.

In order for this thereom to give non-trivial factorizations [tex]n+1[/tex] must not be prime. If this is true and (3) holds then [tex]N[/tex] can be factored easily.

If one thinks carefully about the algorithm for polynomial multiplication you can see that numbers defined by (3) above are quite special with respect to the base [tex]a[/tex].

In order to grasp the full implications of this theorem we need to think about two common algorithms.

1) The first is the base [tex]a[/tex] expansion algorithm used to write a natural number as a sum of powers of a. For example the number [tex]14[/tex] can be written as a sum of powers of [tex]3[/tex] as [tex]14 = 3^2+3+2[/tex]. In general this algorithm can be used to write

(4) [tex] N_a = c_k a^k + c_{k-1} a^{k-1} + ... + c_1 a + c_0[/tex]

Where some of the [tex]c_j[/tex] may be zero and all are less than [tex]a[/tex]. I will denote this algorithm by the capital letter [tex]\beta[/tex] so that [tex]\beta (N,a)[/tex] is equivalent to (4) above.

2) The second algorithm is polynomial multiplication. Now this algorithm proceeds by steps. We first multiply every term by every other term, then we collect terms. Denote this by [tex]P[/tex]. The difference between the result of this algorithm and the base [tex]a[/tex] expansion algorithm is best understood by looking closely at a third related algorithm.

3) The multiplication algorithm for natural numbers is really equivalent to polynomial multiplication taken one step further by carrying so that none of the digits are greater than the base and then evaluation of the resulting polynomial at the base value [tex]a[/tex]. Denote this carry step by [tex]C[/tex].

It follows that if [tex]N = ST[/tex] then [tex]\beta (N,a) = C(P(\beta (S,a),\beta (T,a)))[/tex] and under certain special conditions [tex]\beta(N,a)=P(\beta(S,a),\beta(T,a))[/tex]. One of these special conditions is the thereom stated above. This happens because of two things. First all coefficients of the summation terms on the right are equal to one in the base and the powers that do appear do not cause any duplication of powers, thus there is no carrying.

In the posts that follow I will try to highlight the implications of this. One of the first is that though it would appear that we are looking at a very restricted class of factorizations, in reality a shift operator can often be applied to [tex]\beta(N,a)[/tex] to arrive at a new base where coefficients are all one and the theorem applies. This is typically possible when [tex]\beta(N,a)[/tex] is not the result of carrying, or more formally there exists a pair of factors of [tex]N[/tex] such that the base [tex]a[/tex] expansion and subsequent multiplication produces a polynomial in [tex]a[/tex] with precisely the same coefficients as the base [tex]a[/tex] expansion of [tex]N[/tex].

Clearly this all plays around in an uncomfortable way with our notions of variables and constants. However if you look at it carefully you will see that there is merit here. Furthermore there exist shift operators that can transform certain base [tex]a[/tex] expansions into the form given in (1) above. I will talk about those next.

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