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  • It was proved in 2012/2013: https://en.wikipedia.org/wiki/Goldbach's_weak_conjecture Paul
    paulunderwood@... Feb 24
  • {tst(n,k)=n%(2*k)!=1&&n%(2*k)!=2*k-1&&gcd(k,n)==1&& Mod(Mod(1,n)*(x+1),x^k+1)^n==x^(n%(2*k))+1} This base x+1 test seems to work for k>3. Paul - still testing for pseudoprimality
    paulunderwood@... Feb 16
  • {tst(n,k,a)=gcd(k*(a^3-a),n)==1&&n%(2*k)!=1&& Mod(a,n)^n==a&&Mod(Mod(1,n)*(a+x),x^k+1)^n==a+x^(n%(2*k))} [n,k,a] = [5173601, 3, 5173599] a counterexample and Feitsma's 2^64 2-psp list is strewn with such "x-2" counterexamples. However, it seems for k>3 there are none. Also, for k=4, I have searched all n< 3*10^6 without finding a single counterexample with the "Mod(a,n)^n==a...
    paulunderwood@... Feb 7
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  • {tst(n,k,a)=gcd(k*(a^3-a),n)==1&&n%(2*k)!=1&& Mod(a,n)^n==a&&Mod(Mod(1,n)*(a+x),x^k+1)^n==a+x^(n%(2*k))} [n,k,a]=[1037623, 2, 546] is a counterexample. Paul
    paulunderwood@... Feb 4
  • tst(44801, 3, 229) was the counterexample to the previous test, but Mod(229,n)^n!=229. So here is the wriggled test: {tst(n,k,a)=gcd(k*(a^3-a),n)==1&&n%(2*k)!=1&& Mod(a,n)^n==a&&Mod(Mod(1,n)*(a+x),x^k+1)^n==a+x^(n%(2*k))} (Remember I can include the pre-test Mod(b,n)^n==b; where b = a^k + (-1)^k.) I am running various tests: A straight test on N, for k=2, but this is slow; And for...
    paulunderwood@... Feb 3
  • Please disregard the previous "test" as tst(44801, 3, 229) fools it. Back to the drawing board -- trying to find a way to show that "(x+2)" works for all k, as specified on my UTM bio page Paul -- subject to being hasty
    paulunderwood@... Feb 3
  • I have been making some more tests -- but being in triple loop hell and performing umpteen selfridges per test you can understand that progress is slow. I now have another test (with k>2): {tst(n,k,a)=k>2&&gcd(k,n)==1&&gcd(a^3-a,n)==1&& n%(2*k)!=1&&Mod(Mod(1,n)*(a+x),x^k+1)^n==a+x^(n%(2*k))} Paul -- looking for pseudoprimes
    paulunderwood@... Feb 3
  • I have a great test now and to me it looks like it is fool proof: tst(n,k,a)=gcd(a^3-a,k*n)==1&&gcd(k,n)==1&& n%(2*k)!=1&&Mod(Mod(1,n)*(a+x),x^k+1)^n==a+x^(n%(2*k)) Remember I can pre-test with Mod(b,n)^n==b, where b=a^k+(-1)^k Paul
    paulunderwood@... Feb 2
  • tstq(n,k) = gcd(k,n)==1 && Mod(Mod(1,n)*(2+x),x^k+1)^n==2+x^(n%(2*k)) When verifying this test, for speed, I can pre-test for the implied condition: Mod(b,n)^n==b where b = 2^k+(-1)^k Paul
    paulunderwood@... Feb 1
  • I made another slip the third test should be: tstq(n,k) = gcd(k,n)==1 && Mod(Mod(1,n)*(2+x),x^k+1)^n==2+x^(n%(2*k)) Paul
    paulunderwood@... Jan 23