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Cryptography

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  • $\begingroup$ Thanks for the great explanation. Can you comment on the choice of the polynomial chosen for the modulo? It seems rather odd. Specifically going by what you said, it can be any polynomial as long as the degree is 8. So is there some logic behind the choice? $\endgroup$ Commented Mar 10, 2014 at 6:31
  • 1
    $\begingroup$ First, it cannot be any polynomial of degree $8$. In order to obtain a nice multiplication, it must be an irreducible polynomial of degree $8$. Mathematically, any irreducible polynomial of degree $8$ will do. The choice of this particular polynomial was made for computational reasons, probably in part what you alluded to in 3.: it makes the operation of multiplication by $x$ very easy (and fast) to implement. $\endgroup$ Commented Mar 10, 2014 at 6:37
  • $\begingroup$ It must be a polynomial $p$ with $\deg(p)=8$ and be irreducible. What this means is that it cannot have any (non-trivial) factors. $\endgroup$ Commented Mar 10, 2014 at 9:30
  • $\begingroup$ How do I compute the modulo using a polynomial? For e.g. how to compute - $(x^13 + x^11 + x^9 + x^8 + x^6 + x^5 + x^4 + x^3 + 1) modulo (x^8 + x^4 + x^3 + x + 1)$ $\endgroup$ Commented Mar 10, 2014 at 23:54
  • $\begingroup$ On pencil and paper, just do a long division, but real people use a computer (with e.g. Sage or PARI/GP). ;) $\endgroup$ Commented Mar 10, 2014 at 23:56