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## Latest Messages

Jan 28
A small by-product of research in area of pseudoprimes in k(i): Take a product of two numbers each with shape 4m+3. Let x be this composite number. x is pseudo to base (x-1).Examples 21, 33, 57 etc. (20^20-1)/21 yields a rational integer.

Jan 26
It was not that hard to find this extensive resource regarding tetrahedron. I imagine there will be some follow up soon. <a href="http://www.spilleautomatergratis.org/">www.spilleautomatergratis.org</a>

Jan 26
Let a + ib be a complex number where a and b belong to Z. Then a + ib is a Gaussian integer. We get rational integers if we put b equal to 0. There is atleast one basic difference between rational integers and Gaussian integers. This is illustrated by the following example: 341 is a pseudoprime to base 2 and 23 i.e. (2^340-1)/341 yields a quotient which is a unique rational integer. 21 is a pseudoprime to base (21 + i ). Let ((21+i)^20-1)/21 = x. x is a Gaussian integer; the point is x is also obtained when we change the base to (1-21i), (-21-i) or (-1 + 21i). Hence we do not have a unique base for obtaining x as quotient while applying Fermat's theorem. Incidentally we get the conjugate of x when take base as (1+21i), (21-i),(-21+i) or (-1-21i). In each of the above two cases involving x and its conjugate the four different bases are represented by four points respectively on the complex plane.

Jan 25
Let N = p_1p_2...p_r be an r-factor composite number.If (p_1-1)*(N-1)^(r-2)/(p_2-1)....(p_r-1) is an integer then N is a Devaraj number. All Carmichael numbers are Devaraj numbers but the converse is not true (see A 104016, A104017 and A166290 on OEIS ).

Jan 25
I might have mentioned the following property of polynomials before: let f(x) be a polynomial in x.Then f(x+k*f(x)) is congruent to 0 (mod f(x))(here k belongs to Z. What is new in this message is that it is true even if x is a matrix with elements being rational integers. Also it is true even if x is a matrix with elements being Gaussian integers.

Jan 25
Hi! The coefficients of f(x) belong to Z.

Jan 24
Hi Deva, Speaking of congruence modulo a polynomial means that one considers divisibility in a certain polynomial ring. You don't specify that ring. What is it, i.e. I'm interested what kind of numbers (or others) are the coefficients of f(x)? Jussi

Jan 22
When we take a 3-factor composite such that two are of form 4m+3 and one is of form 4m+1 the said number is a pseudoprime to base the number + i.Example: 3*5*7 = 105. i.e. ((105+i)^104-1)/105 yields a Gaussian integer as quotient.

Jan 22
When we take a composite number, two integers both having shape 4m+3 , we get a composite integer which is pseudo to the base: the number umber + i. Example 21 = 3*7; this number is pseudo to the base (21 + i) i.e. ((21+i)^20-1)/21 is a Gaussian integer. In other words although 21 is not a pseudoprime in k(1), the ring of rational integers, it is pseudo to base (21+i) in the ring of Gaussian integers. This is true of all two prime factor composites(each prime having shape 4m+3). However,this can be verified only if one has pari or similar software in the computer.

Jan 16
11*31 = 341 is a pseudoprime to base 2 and 23. It is also pseudo to base (341 + i). This can be verified only if you have software pari.

[p] hny by pahio Jan 8
Thanks Deva, the same to you! Now it's $5\cdot13\cdot31$.

[p] hny by akdevaraj Jan 8
Happy new year to all! My computer was down; today it has started functioning.Will resume posting messages at the earliest.

Jan 3
In the PM new system, many pictures in the articles have disappeared (e.g. in the article "tractrix"). Today I saw that they may be stored in "Other useful stuff$>$Gallery". How could one transfer such pictures to their pertinent articles?

Jan 3
m and n are integers prove that : sum(i=-m to n) (-1)^|i| * ((2m+i)! (2n-i)!)/((m+i)!^2 (n-i)!^2)) = 1 you can find it here: http://mymathforum.com/math-events/35463-sum.html>