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I need to decompose a large integer (30-40 digits) to an integer unknown with a factorizable polynomial. As a small example: $$ \begin{cases} 119=2x^2+3x \\ x=7 \\ \end{cases} $$

For example:

$$54026707855643784^2+2 \cdot 54026707855643784$$

$$= 2918885161719081869258276809126224$$

This is easy to do for a specific form such as $x^2-y^2 = (x-y)(x+y) $. For instance, if we wanted to find a polynomial of that form for the number $2960$, we could search $x$ such that $x^2-2960$ is a perfect square. We have a lower bound of $\text{ceil}(\sqrt{2960}) = 55$. We soon find that $57^{2}-2960 = 289$, the perfect square of 17, so we get the expression:

$$ \begin{cases} 2960=x^2-289 \\ x=57 \\ \end{cases} $$

However, there is not such an expression for all numbers, such as even very small numbers like $6$. Plus, for larger numbers it could take hundreds of trials till you hit a solution. Thus, I'm trying to find a more general effiicient algorithm for any reducible/factorizable polynomial (making it less likely for flanks like 6 to appear). However, I can find no methods that aren't computationally expensive (e.g. searching a lookup table would take a long time). So is there an algorithm for this?

David G. Stork
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Nirvana
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2 Answers2

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It seems you are seeking so-called algebraic factorizations of integers, i.e. factorizations that arise by representing an integer by reducible polynomials. Your example using a difference of squares representation is known as Fermat's factorization method, which has generalizations such as the quadratic sieve. One can also use various cyclotomic factorizations such as Aurifeuillian factorizations (see the mongraph below for more on such).

John Brillhart, D. H. Lehmer, J. L. Selfridge, Bryant Tuckerman, and S. S. Wagstaff Jr.,
Factorizations of $\,b^n\pm 1,\,$ 2nd ed., Contemporary Mathematics, vol. 22, American Mathematical Society, Providence, RI, 1988.

Bill Dubuque
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You could try representing the integer $z$ in its base-$b$ form for different $b$.

For eg: In base-10,

$$z = 2960 = 2.10^3 + 9.10^2 + 6.10 + 0 = 2960_{10}$$

So,

$$(x, f(x) = (10, 2x^3 + 9x^2 + 6x)$$

In base-$7$, we have $2960 = 11426_7$. So,

$$(x, f(x)) = (7, x^4 + x^3 + 4x^2 + 2x + 6)$$

You can represent $2960$ in many bases $b \in [2,z-1]$ and get different $f(x)$. You can then check if the polynomial is reducible (or irreducible).

See: Methods to see if a polynomial is irreducible

However, this is not any more efficient than factoring $z$. There are faster algorithms for factoring $z$ than this.

vvg
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