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| bio | website | |
|---|---|---|
| location | Vienna, Austria | |
| age | ||
| visits | member for | 1 year, 11 months |
| seen | Apr 13 at 9:03 | |
| stats | profile views | 85 |
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Feb 16 |
comment |
Showing $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$ i think after you have shown that 9=0 (mod p) you are done. you have assumed nothing special about p except that it is a common divisor of n^3+1 and n^2+2. So if you choose p:=gcd(n^3+1,n^2+2) if foolows that gcd is 1,3 or 9 |
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Feb 16 |
comment |
Showing $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$ i think it is not necessary to schow for each of the numbers 1,3,9 that there is an appropriate n. this is not part of the question. if an arbitrary common divisor of n^3+1 and n^2+2 divides 9 than you have shown that gcd(n^3+1,n^2+2) is 1,3 or 9 |
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Feb 16 |
comment |
Showing $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$ gcd(c*a,b)|c*g(a,b), so the only spurious factor that could be added to gcd(n^2+2,2*n-1) by calculating (2*(n^2+1),2*n-1) instead of, is therefore 2. But 2 will not be added because 2*n-1 is odd. |
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Feb 12 |
awarded | Critic |
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Feb 7 |
comment |
RSA calculations what is B? q^3+3*q^2*(1-q)=0.104, if q=0.2 so q_B<q. How do you deduce the probability? |
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Feb 7 |
comment |
Proving a subset is closed under a binary operation on a set as you stated you should prove that a*b in H if a and b in H. a*b in H means that (a*b)*(a*b)=(a*b). can you prove this? |
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Feb 5 |
awarded | Scholar |
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Feb 5 |
accepted | find all positive integers satisfying $2x^2 - y^{14} = 1$ |
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Feb 4 |
awarded | Student |
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Feb 4 |
asked | find all positive integers satisfying $2x^2 - y^{14} = 1$ |
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Feb 4 |
revised |
Can you combine axioms for a vector space? more clear formulation |
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Feb 4 |
revised |
Can you combine axioms for a vector space? more precise, typos |
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Feb 4 |
answered | Can you combine axioms for a vector space? |
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Feb 1 |
revised |
100 Soldiers riddle added 1 characters in body |
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Feb 1 |
comment |
100 Soldiers riddle This is the answer from the original paper @Peter-Shor points to, but "...That means 100 soldiers lost 3 limbs..." is not evident to me. I would prefer the following formulation. A soldier has at most 4 missing limbs. if ther are not more than 9 soldiers that have 4 missing limbs than there are at most 9*4+91*3=309 missing limbs. Therefore there must be at least 10 soldiers that have 4 missing limbs. |
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Feb 1 |
revised |
100 Soldiers riddle added 235 characters in body |
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Feb 1 |
answered | 100 Soldiers riddle |
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Jan 3 |
comment |
An example of an easy to understand undecidable problem there is a typo (a,bba) instead of (a,baa) for tuple 1. but it is not allowed to change only one character |
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Jun 29 |
awarded | Enthusiast |
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Jun 26 |
comment |
Prove that $6|2n^3+3n^2+n$ if you want to use induction then use the fact that for your polynomial $p(n)=n(n+1)(2n+1)$ the following property holds: $$p(n+1)=6(n+1)^2+p(n)$$ |
