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xxx xxx wrote:<br>
<span style="white-space: pre;">><br>
> Lemma:<br>
> p*(-q)=p*q*(-p)<br>
> and respective:<br>
> (-p)*q=p*q*(-q)<br>
> Proof:<br>
> p*(-q)=p*(N-q) - by the data, then<br>
> p*(-q)=p*(p*q-q)=p*pq-p*q=p*q*p-p*q=(p-1)*(p*q)<br>
> (-p)*q=q*(N-p) - by the data, then<br>
> (-p)*q=(p*q-p)*q=p*q*q-p*q=p*q*q-p*q=(q-1)*(p*q)<br>
> Q. E. D.<br>
><br>
><br>
> Like Stanislaw said before be, this Lemma is obvious. You're<br>
> saying that 0=0, and man, this is a thautology!<br>
> You ask why? let N = p*q.<br>
> Then,<br>
> p*q = 0 mod N<br>
> Now, let be -1 the opposite of the unit ( usually called e...)<br>
> 0 = (-1)*0 = (-1)*p*q = (-1*p)*q = (-p)*q<br>
> 0 = 0*(-q) = p*q*(-q)<br>
><br>
> Gypothesis:<br>
> Let N = p*q = A1*B1 + A2*B2... + An*Bn<br>
> Then exists some subset(A1...An) and respective
subset(B1...Bn),<br>
> which<br>
> satisfies for equality:<br>
> A1*(-B1)+A2*(-B2)...+An*(-Bn) = p*(-q)=p*q*(p-1)<br>
> or<br>
> A1*(-B1)+A2*(-B2)...+An*(-Bn) = (-p)*q=p*q*(q-1) <br>
><br>
><br>
> This is another obvious thing!<br>
> if N = sum(A_i*B_i), then<br>
> -N = -1*N = -1*sum(A_i*B_i) = 0 mod N<br>
> and, for the distributive propeties,<br>
> -1*sum(A_i*B_i) = sum (-1*A_i*B_i) = 0 mod N<br>
> <br>
><br>
> If found such (A1...An) and (B1...Bn), we can find p or q by<br>
> dividing<br>
> p*(q-1) on p*q:<br>
> p*(q-1)=p*q*(p-1) => (p*(q-1))/(p*q)=(p-1) => (p-1)+1
= p<br>
> or<br>
> (p-1)*q=p*q*(q-1)=>((-p)*q)/(p*q)=(q-1) => (q-1)+1 = q<br>
><br>
><br>
> Here there's a mistake: p*(q-1) != p*q*(p-1) mod N. in fact, let
N =<br>
> 2*3.<br>
> 2*2 = 4 ! = 6*1 = 0!!!<br>
> Beeing this assumption wrong, all the remaining demostration is<br>
> obviously false...<br>
></span><br>
ok, if your consequences are right, could you disprove this
gypothesis?<br>
<br>
Gypothesis:<br>
Let N = p*q = A1*B1 + A2*B2... + An*Bn<br>
Then exists some subset(A1...An) and respective subset(B1...Bn),
which <br>
satisfies for equality:<br>
A1*B1+A2*B2...+An*Bn = p*q and:<br>
A1*(-B1)+A2*(-B2)...+An*(-Bn) = p*(-q)=p*q*(p-1)<br>
or<br>
A1*(-B1)+A2*(-B2)...+An*(-Bn) = (-p)*q=p*q*(q-1)<br>
<br>
in terms of this gypothesis, could you really prove: there are no one
subsets (A1..An) and respective (B1...Bn) which satisfies equality:<br>
A1*(-B1)+A2*(-B2)...+An*(-Bn) = p*(-q)=p*q*(p-1)<br>
or<br>
A1*(-B1)+A2*(-B2)...+An*(-Bn) = (-p)*q=p*q*(q-1)<br>
?<br>
<br>
Another example in terms of gypothesis:<br>
35 = 2^2*2^3 + 2*1 + 1<br>
then one of possible subsets of 35 is: 4*8 + 2*1 + 1 (4,2,1) and
(8,1,1)<br>
try one of possible cases for test subsets (A1...An) and (B1...Bn):<br>
4*(35-8)+2*(35-1)+1*(-1) = 4*27 + 2*34 + 1*(34) = 108 + 102 = 210<br>
then, 210 / 35 = 6<br>
6+1=7<br>
gcd(35,7)=5<br>
Gypothesis is right (or written above is exception?)<br>
Your sample: 6 = 4 + 2 => 1*4 + 2*1<br>
1*(6-4)+2*(6-1)=12<br>
Divide result by 6: 12/6 = 2<br>
Add one for 2: 2+1 = 3<br>
Test: gcd(6,3)=2<br>
<br>
Any other samples needed?<br>
<br>
More over, while no one present valid proof of incorrectness, it is
correct, right?
<Link.asp?CardId=69;6e;63;6f;72;72;65;63;74;6e;65;73;73;0;4c;69;6e;67;76;6f;55;6e;69;76;65;72;73;61;6c;20;28;45;6e;2d;52;75;29><br>
Has somebody more constructive ideas?<br>
<br>
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