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Factorization will be seen to be just a problem in logic. Factorization of uneven integers having just two prime factors into their two prime factors involves solving for an index, L and two unknowns, M and K in the difference of two squares: M^{2} - K^{2}. Generally, for Q having congruency unity, then Q can be written as:

Q = [4.y + H[1]]^{2} - 2^{2.L}.[4.z + H[2]]^{2}

The optimal Diophantine equation found has solutions, zero for the two unknowns and the coefficients of the unknowns [depending on congruencies] are Y and Z.

All congruencies without exception are modulo [mod] four.

In the case of an uneven integer, Q that is to be factorized, being of the form: [4.k + 1], the integer, M is uneven. The integer, K consists of two factors, 2^{L} which is even and an uneven integral multiplier, so that K is an even integer. In that case, it is better to write: K = 2^{L}.Z, where Z [German: zahlen meaning number] is an uneven integer in this case, so that:

Q = M^{2} - 2^{2.L}.Z^{2}, where Q has the form: [4.k + 1]. M is either [4.m + 1] or [4.m + 3] and Z is either [4.z + 1] or [4.z + 3].

The case:

Q = 2^{2.L}.M^{2} - Z^{2} where Q has the form: [4.k + 3] is a separate case from the above.

The factorization involving binary remainders requires that the interdependence of the integers M, K and L and the given integer, Q to be factorized to be known in advance.

The factorization by deduction procedure [see below] produces an optimal Diophantine equation.

Deductions can be made that start with the integer, Q that make no prior assumptions and require no previous knowledge about the integers, M, K, Z and L apart for the logic of number theory and Diophantine equations [see below].

A lower case x or period [full stop] [.] indicates multiplication. Lower case x is used as the label for the unknown in a quadratic equation below, but the distinction is self evident.

Every variable or numeral in the foregoing is an integer, without exception.

Any uneven composite derived from two prime factors, two composite factors or any combination of a prime and composite factor can be represented as the difference of two squares, [M^{2} - K^{2}]. That is, the difference of two squares property is universal to uneven composites.

There are four classes of classes of composites. The evenness [e] or unevenness [u] of M and K are indicated for each class of classes. Given two factors, q[1] and q[2], where q[2] exceeds q[1], the four classes of classes of composites are: Class of classes [I]: [4.h[1] + 1] x [4.h[2] + 1] giving a Q: [16.h[1].h[2] + 4.h[1] + 4.h[2] + 1]   [M[u], K[e]]

Class of classes [II]:

[4.h[1] + 1] x [4.h[2] + 3] giving a Q: [16.h[1].h[2] + 12.h[1] + 4.h[2] + 3] [M[e], K[u]]

Class of classes [III]:

[4.h[1] + 3] x [4.h[2] + 1] giving a Q: [16.h[1].h[2] + 4.h[1] + 12.h[2] + 3] [M[e], K[u]]

Class of classes [IV]:

[4.h[1] + 3] x [4.h[2] + 3] giving a Q: [16.h[1].h[2] + 12.h[1] + 12.h[2] + 9] [M[u], K[e]]

In particular reference to the heading, Deductions below, it is useful to know in advance the congruencies of M and K.

More generally, we can write:

[i]          M^{2} - K^{2} = 2^{2.L}.Y^{2} - Z^{2}

and:

[ii]         M^{2} - K^{2} = Y^{2} - 2^{2.L}.Z^{2}

In [i], M = 2^{L}.Y and K = Z and in [ii], M = Y and K = 2^{L}.Z In each case, [i] and [ii], Y and Z are uneven integers. This implies that Y and Z can have congruencies of unity and three in every combination. Writing Y = [4.y + 1] or Y = 4.Y + 3] and Z = [4.z + 1] or Z = [4.z + 3] generates all of the combinations [see below].

It was noticed that case [i] only occurs when Q has the form: [4.k + 3] and case [ii] only occurs when Q has the form: [4.k + 1].

The object now is by writing out all of the combinations of congruencies for case [i] and case [ii] [separately] to find which combinations logically equated to the appropriate generalizations of Q comprising the classes: [I}, .. [IV] above result in logically correct equalities and which result in, if any, logical contradictions.

Case [i] Q has the form; [4.k + 3]

From [II]

[j] [4.h[1] + 1].[4.h[2] + 3] = 2^{2.L}.[4.y + 1]^{2} - [4.z + 1]^{2}

[k] [4.h[1] + 1].[4.h[2] + 3] = 2^{2.L}.[4.y + 1]^{2} - [4.z + 3]^{2}

[l] [4.h[1] + 1].[4.h[2] + 3] = 2^{2.L}.[4.y + 3]^{2} - [4.z + 1]^{2}

[m] [4.h[1] + 1].[4.h[2] + 3] = 2^{2.L}.[4.y + 3]^{2} - [4.z + 3]^{2}

From [III]

[n] [4.h[1] + 3].[4.h[2] + 1] = 2^{2.L}.[4.y + 1]^{2} - [4.z + 1]^{2}

[p] [4.h[1] + 3].[4.h[2] + 1] = 2^{2.L}.[4.y + 1]^{2} - [4.z + 3]^{2}

[q] [4.h[1] + 3].[4.h[2] + 1] = 2^{2.L}.[4.y + 3]^{2} - [4.z + 1]^{2}

[r] [4.h[1] + 3].[4.h[2] + 1] = 2^{2.L}.[4.y + 3]^{2} - [4.z + 3]^{2}

From [I]

[ii] [a] [4.h[1] + 1].[4.h[2] + 1] = [4.y + 1]^{2} - 2^{2.L}.[4.z + 1]^{2} [b] [4.h[1] + 1].[4.h[2] + 1] = [4.y + 1]^{2} - 2^{2.L}.[4.z + 3]^{2}

[c] [4.h[1] + 1].[4.h[2] + 1] = [4.y + 3]^{2} - 2^{2.L}.[4.z + 1]^{2}

[d] [4.h[1] + 1].[4.h[2] + 1] = [4.y + 3]^{2} - 2^{2.L}.[4.z + 3]^{2}

From [IV]

[e] [4.h[1] + 3].[4.h[2] + 3] = [4.y + 1]^{2} - 2^{2.L}.[4.z + 1]^{2}

[f] [4.h[1] + 3].[4.h[2] + 3] = [4.y + 1]^{2} - 2^{2.L}.[4.z + 3]^{2}

[g] [4.h[1] + 3].[4.h[2] + 3] = [4.y + 3]^{2} - 2^{2.L}.[4.z + 1]^{2}

[h] [4.h[1] + 3].[4.h[2] + 3] = [4.y + 3]^{2} - 2^{2.L}.[4.z + 3]^{2}

A Strategy For Assigning the Proper Congruencies to Y and Z Above.

Case of Q having congruency unity.

The first and fourth of the four classes, that is [I] and [IV] occur where the congruency of Q is unity. This means that Q has been composed of two factors that are either of the form:

[I] 4.h[1] + 1]x [4.h[2] + 1] or exclusively [IV] [4.h[1] + 3] x [4.h[2] + 3] with h[1] representative of q[1] the smaller of the two factors.

If Q had been derived via [I] and we set Q as if it had been derived from [IV}, then a logically false situation occurs, that of an even integer being equated to an uneven integer.

Example: Q = 65 [5, 13] Note that q[1] = 5 and q[2] = 13 each have congruency unity [1]. Let us equate this to a class [IV] case. That is:

65 = [16.h[1].h[2] + 12.h[1] + 12.h[2] + 9]

The prime factors of sixty-five are alleged to have congruencies each of three.

Simplifying the above expression, we obtain:

56 = 16.h[1] + 12.h[1] + 12.h[2]

Division by four, since this can be done, gives:

14 = 4.h[1].h[2] + 3.h[1] + 3.h[2]

The left side of this equation is even and therefore the right side has to be even as well. The conclusion is that [a] h[1] and h[2] have each got to be even or exclusively [b] h[1] and h[2] have each got to be uneven. Taking case [a] first, set h[1] = 2.k[1] and h[2] = 2.k[2]. Substitution leads to:

14 = 4.[2.k[1]].[2.k[2]] + 3.[2.k[1]] + 3.[2.k[2]]

Simplification results in:

14 = 16.k[1].k[2] + 6.k[1] + 6.k[2]

Division by two gives:

7 = 8.k[1].k[2] + 3.k[1] + 3.k[2]

The conclusion at this juncture is that [c] k[1] has to be even and k[2] has to be uneven or alternatively [d] k[1] has to be uneven and k[1] has to be even.

Instance [c] Let k[1] = 2.m[1] and k[2] = [2.m[2] + 1]. Making these substitutions:

7 = 8.[2.m[1]].[2.m[2] + 1] + 3.[2.m[1]] + 3.[2.m[2] + 1]. Simplifying gives:

7 = 32.m[1].m[2] + 16.m[1] + 6.m[1] + 6.m[2] + 3. That is:

4 = 32.m[1].m[2] + 22.m[1] + 6.m[2]. Division by two gives:

2 = 16.m[1].m[2] + 11.m[1] + 3.m[2]

The conclusion now is that either [e] m[1] and m[2] are each even or exclusively [f] m[1] and m[2] are each uneven. Case [e] Putting m[1] = 2.n[1] and m[2] = 2.n[2] and substituting:

2 = 16.[2.n[1].2.n[2]] + 11.[2.n[1]] + 3.[2.n[2]]. Simplifying:

2 = 64.n[1].n[2] + 22.n[1] + 6.n[2]. Division by two gives:

1 = 32.n[1].n[2] + 11.n[1] + 3.n[2]

The conclusion is that either [g] n[1] is even and n[2] is uneven or [h] n[1] is uneven and n[2] is even. Taking case [g], let n[1] = 2.p[1] and n[2] = 2.p[2] + 1. [no connection with prime p]. Simplifying:

1 = 32.[2.p[1].[2.p[2] + 1]] + 11.[2.p[1]] + 3[2.p[2] + 1]] Simplifying:

1 = 128.p[1].p[2] + 64.p[1] + 22.p[1] + 6.p[2] + 3. Simplifying:

- 2 = 128.p[1].p[2] + 86.p[1] + 6.p[2]. Division by two gives:

- 1 = 64.p[1].p[2] + 43.p[1] + 3.p[2]

By hypothesis, h[1] and h[2] are positive. Therefore k[1] and k[2] are positive. Therefore k[1] and k[2] are positive. Therefore m[1] and m[2] are positive. Therefore n[1] and n[2] are each positive. Therefore p[1] and p[2] are each positive. In the above equation, the left side is intrinsically negative and the right side is positive by hypothesis. This is a logical contradiction and therefore the original hypothesis that Q = 65 was derived from case [IV] was and remains false. The conclusion is that without the labor of going through every combination of substitutions we can distinguish case [I] from case [IV].

The congruency strategy for Y and Z.

This having been done, taking a value of Q that follows case [I]. There are four combinations of congruencies of Y and Z. Setting Y = [4.y + 1], Z = 4.Z + 1] etc. Rewriting out the above [a] through to [d], case[I], but multiplying out the left side, we have:

[a] [16.h[1].h[2] + 4.h[1] + 4.h[2] + 1] = [4.y + 1]^{2} - 2^{2.L}.[4.z + 1]^{2}

[b] [16.h[1].h[2] + 4.h[1] + 4.h[2] + 1] = [4.y + 1]^{2} - 2^{2.L}.[4.z + 3]^{2}

[c] [16.h[1].h[2] + 4.h[1] + 4.h[2] + 1] = [4.y + 3]^{2} - 2^{2.L}.[4.z + 1]^{2}

[d] [16.h[1].h[2] + 4.h[1] + 4.h[2] + 1] = [4.y + 3]^{2} - 2^{2.L}.[4.z + 3}^{2}

These are four of the sixteen cases above. Since the right sides of [a] to [d] are each distinct, there has to be in each of the four cases distinct properties of h[1] and h[2] that will characterize the congruencies of Y and Z. This has to be implicit in the integer Q. Simplifying [a]:

16.h[1].h[2] + 4.h[1] + 4.h[2] = 16.y^{2} + 8.y - 2^{2.L}.[4.z + 1]^{2}

Dividing each side by four:

4.h[1].h[2] + h[1] + h[2] = 4.y^{2} + 2.y - 2^{2.L - 2}.[4.z + 1]^{2}

The conclusion is that [i] h[1] and h[2] are each even, or exclusively [ii] h[1] and h[2] are each uneven. The question is: working through the four cases [a] to [d] is this sufficient to establish the congruencies of Y and Z, or is it better to continue with substitutions for h[1] and h[2] using the two choices [i] and [ii] and serially process [a] through to [d] ? The best way in the present situation is to process [a] through to [d] in parallel until it can be certain that the four cases are distinct with respect to h[1] and h[2] and the respective congruencies.

Simplifying [b]: ...

The integer, L takes values: selected from: [1, 2, 3 ... ]. The integers z and x take values selected from: [0, 1, 2, 3, ... ]

Some of the cases [a] through to [r] may be invalid, some are identical to others. A comprehensive tabulation should reduce to number of cases and establish the template or pattern.

There are sixteen ways of combining h[1], h[2] and y, z as shown above. This means that for a fixed L and z, given a sequence of x, then this produces a sequence of composites, Q. All together there are: 16.L.z sequences of composites, Q and 16.L.z associated sequences of ordinals of Q. Some of these sequences might be indistinguishable [degeneracy] giving perhaps only 8.L.z or 4.L.z distinct sequences or even only 2.L.z sequences. Any definite composite composed of two distinct factors irrespective of each being prime, composite or both would be included in one of the [16.L.z] sequences, there being no excluded composites. Perfect squares where h[1] and h[2] are identical are the exception. Some numerical tabulations commencing not with h[1] and h[2] but with fixed values of L and z and a progressive of x would make matters clearer and elucidate the nature of the sequences. Then given a definite numerical composite, the appropriate combination from [a], ... [r] could be identified and the composite factored by deduction as described in: Factorization/Integer/Binary remainders. Composites having three or more prime factors would appear as composites with two factors, one or both of these factors being a composite.

We have equated these to cases [I] and [II] of the generalizations of Q, the integer to be factorized expressed in terms of h[1] and h[2], above.

The first two members of the class of classes [I] are: [[5, 9], [5, 13], ... ], ... [[9, 13], [9, 17], ... ], [[13, 17], [13, 21] ... ] ... etc. for the other three classes, with five being the smallest member of the [4.h + 1] set and three being the smallest member of the [4.h + 3] set of integers. The degenerate case of unity as the factor, q[1] was left out.

Examples of factorization by solving remainders and an index, one from each class are:

[1] Q[6] = 65 [5, 13], [2] Q[8] = 323 [17, 19], [3] Q[8] = 319 [11, 29] and [4] Q[9] = 713 [23, 31]. An interesting easily proved result about uneven integers that can be factorized.

Given Q, then we can write: Q = M^{2} - K^{2}

= [M - K].[M + K]

The two prime, composite or mixed factors are: M - K and M + K.

Proof: The interval between two successive uneven integers is known to be two. Taking q[2] to be greater than q[1], that is:

[q[2] - q[1]] = 2.K

multiplying through by q[2]:

q[2]^{2} - q[1].q[2] = 2.K.q[2]

Given that Q = q[1].q[2] by definition, this becomes:

q[2]^{2} - Q = 2.K.q[2]

This is a quadratic in q[2],

x^{2} - 2.K.x + [- Q] = 0

The solutions are x[1] = K + [K^[2} + Q]^{1/2} and x[2] = K - [K^{2} + Q]^{1/2}

The discriminant D = [K^{2} + Q] has to be a perfect square, say M^[2} for q[1] and q[2] to be integers.

Then M^{2} = [K^{2} + Q].

The final term in the quadratic: a.x^{2} + b.x + c = 0 is the product of the roots, x[1] and x[2].

That is: x[1].x[2] = c

The result then follows: [- Q] = x[1].x[2] or

[- Q] = [K^{2} - M^{2}]

or

Q = M^{2} - K^{2} and that is:

Q = [M - K].[M + K], as above.

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Deductions.

Every congruency in the list below is modulo [mod] four.

A List of Numerals to set out so as to be able to recognize the congruencies of X[1] and X[2] with only Q as the datum [see below for a definition of each numeral].

The list is: [Q, g[0], J[0], q[1], h[1], H[1], q[2], h[2], H[2], X[1], N[1], X[2], N[2], L[j]]

The congruencies of X[1] and X[2] are respectively: N[1] and N[2]. The index of two is L[j] and the subscript [j] is [1, 2] one or two.

The integer to be factored, Q can be written: Q = 4.g[0] + J[0]. The integer g[0] can be written: g[0] = 2^{K[0]}.[4.g[1] + J[1]] etc.

The ordinal of Q is n and: n = [1 + Q]/2 and has an easily distinguished congruency unity or three. The prime factors q[1] and q[2] have respective congruencies, H[1] and H[2].

In terms of h[1], H[1], h[2] and H[2], q[1] = 4.h[1] + H[1] and q[2] = 4.h[2] + H[2]

In terms of M and K, if Q is of congruency unity,

q[1] = X[1] - 2^{L[2]}.X[2]

and

q[2] = X[1] + 2^{L[2]}.X[2]

X[1] = [4.x[1] + N[1]] and X[2] = [4.x[2] + N[2]]

Explicitly, q[1] = [4.x[1] + N[1]] - 2^{L[2]}.[4.x[2] + N[2]] and

q[2] = [4.x[1] + N[1]] + 2^{L[2]}.[4.x[2] + N[2]]

In terms of M and K, if Q is of congruency three,

q[1] = 2^{L[1]}.X[1] - X[2]

and

q[2] = 2^{L[1]}.[X[1]] + X[2]

X[1] = [4.x[1] + N[1]] and X[2] = [4.x[2] + N[2]]

Explicitly:

q[1] = 2^{L[1]}.[4.x[1] + N[1]] - [4.x[2] + N[2]]

q[2] = 2^{L[1]}.[4.x[1] + N[1]] + [4.x[2] + N[2]]

The designation X[1] replaces Y and the designation X[2] replaces Z, above. Index L is replaced by L[1]which is associated with X[1] or L[2] which is associated with X[2], having respective congruencies, N[1] and N[2]. The integer [j] of L[j] takes values: j = 1 or j = 2.

Generally,

If H[1] is the same as H[2], each say H[1], then Q has congruency unity [mod] four and:

[4.h[1] + H[1]].[4.h[2] + H[1]] = [4.x[1] + N[1]]^{2} - 2^{2.L[2]}.[4.x[2] + N[2]]^{2}

There are two cases.

If H[1] and H[2] are different, then Q has congruency three and:

[4.h[1] + H[1]].[4.h[2] + H[2]] = [2^{2.L[1]}.[4.x[1] + N[1]]^{2} - 4.x[2] + N[2]]^{2}

There are two cases.

The purpose of tabulating the list above is to find the pattern of N[1] and N{2] and less immediately, the pattern of L[1] or L[2].

The matter of assigning the proper congruencies to X{1] and[X2], otherwise labeled Y and Z respectively is being treated as a theoretical problem above.

.................

Some logic: given that U and u are uneven class integers and E and e are even class integers, then:

U = e + u, E = u + u, E = e + e U = u and E = e are logically true, but U = e and E = u are logically false or contradictory. The logical equations for multiplication are: E = e.e, E = e.u and U = u.u. The equation, 2^{K}.u = E is logically true provided that the positive integer, K is equal to unity or greater. If K = 0 or negative, then this equation is logically false.

..........

The smallest composite, Q having distinct factors and congruency, unity is Q = 21 [3, 7]. The smallest composite, Q having distinct factors and congruency, three is Q = 15 [3, 5]. The squares of uneven integers all have congruencies of unity. An example is Q = 49 [7, 7] has congruency unity.

Example [1]: factorization of Q[6] = 65 [5, 13]

Sixty-five is congruent to unity, modulo four, or alternatively: [65 - 1]/4 = 16, remainder zero. That is, M^{2} is uneven and K^{2} is even and this is always found to be the case for Q congruent to unity, modulo four. M = 9 and K = 4.1 That is K = 2{L}.[4.k + 1], presenting the congruency of K.

By trial and error, M has the form [4.m + 1] and K has the form 2^{L}.[4.k + 1]. There are only four combinations to try, and one of them is correct.

We have:

65 = [16.m^{2} + 8.m + 1] - 2^{2.L}.[16.k^{2} + 8.k + 1]

64 = [16.m^{2} + 8.m] - 2^{2.L}.[16.k^{2} + 8.k + 1]

The least value of L is unity, one, to make K even. We can therefore divide the above equation by four [without producing fractions] giving:

16 = 4.m^{2} + 2.m - 2^{2.L - 2}.[16k^{2} + 8.k + 1]

If [2.L - 2] is zero, then the left side being even and the right side containing unity along with the even terms, 4.m^{2},2.m, 4.k^{2} and 8.k would be a contradiction.

Dividing by two gives:

8 = 2.m^{2} + m - 2^{2.L - 3}.[16.k^{2} + 8.k + 1]

The term: 2^{2.L - 3} can only be even because [2.L - [2.J - 1]] is uneven, J being unity in this term. Therefore m is even. Let m = 2.h substitution for m gives:

8 = 2.[2.h]^{2} + 2.h - 2^{2.L - 3}.[16h^{2} + 8.h + 1]

Division by two gives:

4 = 4.h^{2} + h - 2^{2.L - 4}.[16.k^{2}+ 8.k + 1]

For a solution, the index [2.L - 2.J] has to be zero, taking the least value of J. The above equation satisfies that condition with J = 2. If the value, J = 3 was taken, 2^{2 - 6} would involve a factor of 1/16. Clearing this factor would then lead to a logical contradiction, an even integer equal to an uneven integer. Therefore L = 2. Rewriting the above equation:

4 = 4.h^{2} + h - 16k^{2} - 8.k - 1

This equation has an even left side. The right side contains unity, an uneven integer and the unknown, h having an uneven coefficient, unity. That is 1.h.

To avoid a logical contradiction, the unknown h has to be an uneven integer. Let us put:

h = 2.Y + 1. Substitution into the above equation results in:

4 = 4.[2.Y + 1]^{2} + [2.Y + 1] - 16.k^{2} - 8.k - 1

Simplifying in stages:

4 = 4.[4.Y^{2} + 4.Y + + 1] + [2.Y + 1] - 16.k^{2} - 8.k - 1

4 = 16.Y^{2} + 16.Y + 4 + 2.Y + 1 - 16.k^{2} - 8.k - 1

That is:

0 = 16.Y^{2} + 18.Y - 16.k^{2} - 8.k

Dividing by two finally produces the Diophantine equation:

0 = 8.Y^{2} + 9.Y - 8.k^{2} - 4.k

The immediately found solutions are: Y = 0 and k = 0 Therefore h = 1 Integer, m = 2.h and therefore m = 2. M = 9 Therefore K = 4 since k is zero.

As in Example [2] below, the coefficients of Y and k in some way respectively represent M and K and in this example [1], there is equality.

Alternatively, reversing the substitutions: h = 2.Y + 1 and m = 2.h makes m = 2 and with k = 0 and L = 2, the prime factors are found.

...............................................................................................

Example [3]: factorize: Q = 742529 [733, 1013] [M = 873 and K = 140. The integer, K can be written as K = 2^{2}.35] Only the congruencies of M and K are known in advance. One stage is done at a time primarily to avoid mistakes. The final result of the logic is a Diophantine equation in two unknowns.

Solution: Q has the form: [4.k + 1]. Therefore M is uneven and K is even.

By trial and error or by further research, M = [4.m + 1] and K = 2^{L}.[4.z + 3]

Thus:

742529 = [16.m^{2} + 8.m + 1] - 2^{2.L}.[16.z^{2} + 24.z + 9]

742528 = 16.m^{2} + 8.m - 2^{2.L}.[16.z^{2} + 24.z + 9]

We can divide the above equation by four because the minimum value of the index L is unity.

185632 = 4.m^{2} + 2.m - 2^{2.L - 2}.[16.z^{2} + 24.z + 9]

If [2.L - 2] is zero, then the above equation would have an even left side and an uneven right side. This is logically inconsistent and therefore [2.L - 2] is a positive even integer and we can divide by two.

92816 = 2.m^{2} + m - 2^{2.L - 3}.[16.z^{2} + 24.z + 9]

The index: [2.L - 3] cannot be zero because L is an integer. The conclusion is that m is an even integer. Let m = 2.h Substitution for m gives:

92816 = 2.[2.h]^{2} + [2.h] - 2^{2.L - 3}.[16.z^{2} + 24.z + 9]

This equation after division by two, becomes:

46408 = 4.h^{2} + h - 2^{2.L - 4}.[16.z^{2} + 24.z + 9]

This equation is the first to satisfy the condition that h is uneven and [2.L - 4] = 0 Therefore L = 2.

Rewriting this equation:

46408 = 4.h^{2} + h - 16.z^[2} - 24.z - 9

Putting h = 2.j + 1 and substituting

Transferring the nine to the left side of the above equation and noting the minus sign, we find:

46417 = 4.[2.j + 1]^{2} + [2.j + 1] - 16.z^{2} - 24.z

This is:

46417 = 4.[4.j^{2} + 4.j + 1] + [2.j + 1] - 16.z^{2} - 24.z

Simplifying:

46417 = 16.j^{2} + 16.j + 4 + 2.j + 1 - 16.z^{2} - 24.z

This becomes:

46417 = 16.j^{2} + 18.j + 5 - 16.z^{2} - 24.z

Transferring the five to the left side of the above equation gives:

46412 = 16.j^{2} + 18.j - 16.z^{2} - 24.z

Dividing each side of the above by two gives:

23206 = 8.j^{2} + 9.j - 8.z^{2} - 12.z

In the term: 9.j, the integer j has to be even because the left side of this equation is even.

Let j = 2.g. Substitution gives:

23206 = 8.[2.g]^{2} + 9.[2.g] - 8.z^{2} - 12.z

That is:

23206 = 32.g^{2} + 18.g - 8.z^{2} - 12.z

Division by two gives:

11603 = 16.g^{2} + 9.g - 4.z^{2} - 6.z

Because the left side is uneven, g has to be uneven as well. Let g = 2.X + 1 Substitution for g gives:

11603 = 16.[2.X + 1]^{2} + 9.[2.X + 1] - 4.z^{2} - 6.z

This becomes:

11603 = 64.X^{2} + 64.X + 16 + 18.X + 9 - 4.z^[2} - 6.z

Transferring twenty-five to the left side of the equation gives:

11578 = 64.X^{2} + 82.X - 4.z^{2} - 6.z

Dividing by two gives:

5789 = 32.X^{2} + 41.X - 2.z^{2} - 3.z

In the above equation, we can substitute X = 2.Y + 1 and z = 2.H leaving one uneven integer on the right side of the equation. Substitution gives:

5789 = 32.[4.Y^{2} + 4.Y + 1] + 41.[2.Y + 1] - 2.[2.H]^{2} - 3.[2.H]

Simplifying:

5789 = 128.Y^{2} + 128.Y + 32 + 82.Y + 41 - 8.H^{2} - 6.H

Simplifying gives:

5716 = 128.Y^{2} + 210.Y - 8.H^{2} - 6.H

Division by two gives:

2858 = 64.Y^{2} + 105.Y - 4.H^{2} - 3.H

Because the left side is even, [i] Y and H have each got to be even or alternatively,[ii] each have to be uneven.

Trying alternative [i], Set Y = 2.a and H = 2.b Substitution gives:

2858 = 256.a^{2} + 210.a - 16.b^{2} - 6.b

Division by two gives:

1429 = 128.a^{2} + 105.a - 8.b^{2} - 3.b

Either a is uneven and b is even or exclusively a is even and b is uneven.

Let a = 2.W + 1 and b = 2.Z [upper case] Substitution gives;

1429 = 128.[4.W^{2} + 4.W + 1] + 105.[2.W + 1] - 8.[2.Z]^{2} - 3.[2.Z]

Simplifying:

1429 = 512.W^{2} + 512.W + 128 + 210.W + 105 - 32.Z^{2} - 6.Z

1196 = 512.W^{2} + 722.W - 32.Z^{2} - 6.Z

Division by two gives:

598 = 256.W^{2} + 361.W - 16.Z^{2} - 3.Z

The above is a Diophantine equation with solutions: W = 1 and Z = 1

Check: 598 = 256 + 361 - 16 - 3                  [this is correct]

Reversing the substitutions gives: a = 3 and b = 2. Now Y = 2.a and H = 2.b. This gives:

Y = 6 and H = 4 Next X = 2.Y + 1 and z = 2.H [lower case z] This gives: X = 13 and z = 8 Next: g = 2.X + 1

This gives: g = 27 The substitution; j = 2.g gives: j = 54 Next: h = 2.j + 1 This giving:

h = 109 Almost there, m = 2.h, so that m = 218 Finally, M = [4.m + 1]

That is: M = 873   [This is correct]

K = 2^{L}.[4.z + 3] Substitution for the known values, L = 2 and z = 8 gives:

K = 140            [This is correct]

But checking: q[1] = M - K and q[2] = M + K gives

q[1] = 873 - 140 = 733

q[2] = 873 + 140 = 1013.

The above Diophantine equation that along with the index, L returned the correct value of M and K is not optimally Diophantine. Rewriting this equation:

598 = 256.W^{2} + 361.W - 16.Z^{2} - 3.Z

The left side of this equation is an even integer, 598. The logical conclusion is that [a] W and Z have to be each even, or the alternative is that [b] W and Z have to be each uneven. In example [2], Q = 2701, the uneven choice was used, but the even choice, while resulting in a non Diophantine equation still gave the correct answers. Taking then alternative [b], set:

W = 2.u + 1 and Z = 2.v + 1 Using these substitutions, the equation becomes:

598 = 256.[4.u^{2} + 4.u + 1] + 361.[2.u + 1] - 16.[4.v^{2} + 4.v + 1] - 3.[2.v + 1]

Simplifying:

598 = 1024.u^{2} + 1024.u + 256 + 722.u + 361 - 64.v^{2} - 64.v - 16 - 6.v - 3

This becomes:

0 = 1024.u^[2} + 1746.u - 64.v^{2} - 70.v

Division by two presents the optimal Diophantine equation:

0 = 512.u^{2} + 873.u - 32.v^{2} - 35.v

The obvious solutions are: u = 0 and v = 0. Notice the coefficients of u and v and compare with the known values of X[1] and X[2] of M and K.

In the optimal Diophantine equation, the left side is even, zero [refer to the logic above if not sure about zero being even]. The right side contains the terms: + 873.u and - 35.v. This means that we can substitute, say [a] u = 2.c + 1 and v = 2.d + 1,  or alternatively [b] u = 2.c and v = 2.d The resulting equation would probably return the correct answers, but is past optimality. This iteration of substitutions could be continued indefinitely.

The above are the found prime factors of 742529. The square root of 742529 is approximately 861.7 The classical factorization solution would require many long divisions by three, two and one digit decimal prime numbers. How many? Counting from a table of primes that have been printed in a regular geometrical array, one hundred and forty-eight long and short decimal prime divisions [excluding the prime two [2] which properly is in a different class]. How many divisions by two in this deductive solution? The answer is that there are nine successive divisions by two to arrive at the prime factors of 742529 via M and K. The deductive solution to factorization is much more intricate than the classical solution.

Referring to example [2], apparently, there was a crisis in the progress of the solution in this Example [3], but the proper option had been chosen and as a result, the crisis was not noticed. This may have been purely by chance with a fifty per cent probability [much better than a lottery]. This looks like the equation: 5789 = 32.X^{2} + 41.X - 2.z^{2} - 3.z We could have substituted X = 2.Y and z = 2.H + 1 in place of what had been substituted, which was the opposite substitution.

The only assumption was the congruencies of M and K found by knowing the factors in advance. Research should make this one in four choice that is a twenty-five per cent probability unnecessary.

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Example [4]: factorize Q = 51755383 [6529, 7927] by deduction. Solution: The square root of Q is 7194.12... approx. Finding the prime factors by the classical procedure you would need about nine hundred long decimal divisions. This example might need too many instances of substitutions to do by hand.

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Fifteen [15] is the smallest composite having a congruency of three and distinct prime factors. Twenty-one [21] is the smallest composite having a congruency of unity and two distinct prime factors. The squares of integers have a congruency of unity irrespective of whether the integer is a prime or not. Examples of this are forty-nine, Q = 49 [7, 7] and eighty-one, Q = 81 [9, 9]. ...............................................................................................

Example [5]: Factorize Q = 15 [3, 5] [M = 4, K = 1, L[1]/L = 2, X[1]/Y = 1, X[2]/Z = 1]

Solution: the starting equation is:

15 = 2^{2.L}.[16.y^{2} + 8.y + 1] - [16.z^{2} + 8.z + 1]SHAWWPG19410425 (talk) 15:01, 19 March 2014 (UTC)SHAWWPG19410425 (talk) 09:43, 17 March 2014 (UTC)

SHAWWPG19410425 (talk) 13:08, 15 March 2014 (UTC) SHAWWPG19410425 (talk) 00:58, 15 March 2014 (UTC) SHAWWPG19410425 (talk) 00:47, 15 March 2014 (UTC)SHAWWPG19410425 (talk) 10:10, 14 March 2014 (UTC)SHAWWPG19410425 (talk) 18:08, 11 March 2014 (UTC) SHAWWPG19410425 (talk) 12:00, 11 March 2014 (UTC)SHAWWPG19410425 (talk) 13:16, 9 March 2014 (UTC) SHAWWPG19410425 (talk) 13:11, 7 March 2014 (UTC)

Source: Mathematics contains its own verification. There would be no point in copying the work of someone else. In any case, the above is no where to be found in the established literature.SHAWWPG19410425 (talk) 16:52, 11 March 2014 (UTC)

SHAWWPG19410425 (talk) 21:18, 5 March 2014 (UTC)SHAWWPG19410425 (talk) 23:21, 4 March 2014 (UTC) SHAWWPG19410425 (talk) 10:27, 19 February 2014 (UTC)]SHAWWPG19410425 (talk) 15:25, 15 February 2014 (UTC) SHAWWPG19410425 (talk) 13:12, 8 February 2014 (UTC)SHAWWPG19410425 (talk) 13:33, 28 May 2012 (UTC)SHAWWPG19410425 (talk) 06:55, 9 February 2014 (UTC)