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January 23

Given an integer for which a square root against a large semiprime modulus exists, is the number of possible square roots always 4 ?

Simple question : take a large semiprime $n$ . Take an integer $i$ such as √i%n has an existing solution. Is the number of possible solutions always 4 in such a case ? (or 2 if the modular inverse are excluded) 2A01:E0A:401:A7C0:E4AA:FB65:CDCC:FA58 (talk) 11:09, 23 January 2025 (UTC)[reply]

Yes, because 1 always has four square roots modulo an (odd) semiprime. Tito Omburo (talk) 12:11, 23 January 2025 (UTC)[reply]

January 28

Fermat polygonal number theorem for centered polygonal numbers and generalized polygonal numbers

Fermat polygonal number theorem is that every positive integer is a sum of <=n n-gonal numbers (n such numbers seems to not be always needed, e.g. only needed 4 for n=6, so what is the smallest m such that every positive integer is a sum of <=m n-gonal numbers? I only know that m<=n), but what about centered n-gonal numbers and generalized n-gonal numbers (e.g. OEIS: A001318 for n=5), what is smallest number m such that every positive integer is a sum of <=m such numbers? Also, what about n-dimensional simplex numbers and n-dimensional cross-polytope numbers (generalization of Pollock's conjectures to higher dimension)? (For n-dimensional hypercube, there is already Waring's problem) 220.132.216.52 (talk) 12:09, 28 January 2025 (UTC)[reply]

January 29

Example of Jones formula

If we let the 26 variables be a=1, b=2, c=3, …, z=26, then what are the solutions of the set of 14 Diophantine equations? 118.170.15.127 (talk) 11:16, 29 January 2025 (UTC)[reply]

The first equation of the set of 14 equations, is:

wz+h+j-q=0.

Using the assignment of values

w=23,z=26,h=8,j=10,q=17,

this becomes

599=0.

Seeing as this ain't so, the system has no solutions.

If the set had a solution under this specific assigment, it would be this:

a=1,b=2,c=3,...,z=26.

The question is a bit like, "letting

x=1,

solve the equation

2x=4.

" --Lambiam 12:28, 29 January 2025 (UTC)[reply]

You can't just take a random set of values and solve for x - there is no x in the usual sense of single variable polynomial in the formula - one must think of all a to z as 26 different x's. What the Jones formula does is provide one with a way of proving a number is prime by supplying 26 numbers and showing the result of that formula is the prime number. Which is quite amazing - one just needs to do a small constant number of operations - addition subtraction multiplication and comparisons with zero. However the numbers can be of the order of the prime to the power of itself - so definitely not practical to generate never mind use! NadVolum (talk) 13:09, 29 January 2025 (UTC)[reply]

There is a prime generating polynomial of Jones formula:

{\begin{aligned}&(k+2)(1-{}\\[6pt]&[wz+h+j-q]^{2}-{}\\[6pt]&[(gk+2g+k+1)(h+j)+h-z]^{2}-{}\\[6pt]&[16(k+1)^{3}(k+2)(n+1)^{2}+1-f^{2}]^{2}-{}\\[6pt]&[2n+p+q+z-e]^{2}-{}\\[6pt]&[e^{3}(e+2)(a+1)^{2}+1-o^{2}]^{2}-{}\\[6pt]&[(a^{2}-1)y^{2}+1-x^{2}]^{2}-{}\\[6pt]&[16r^{2}y^{4}(a^{2}-1)+1-u^{2}]^{2}-{}\\[6pt]&[n+\ell +v-y]^{2}-{}\\[6pt]&[(a^{2}-1)\ell ^{2}+1-m^{2}]^{2}-{}\\[6pt]&[ai+k+1-\ell -i]^{2}-{}\\[6pt]&[((a+u^{2}(u^{2}-a))^{2}-1)(n+4dy)^{2}+1-(x+cu)^{2}]^{2}-{}\\[6pt]&[p+\ell (a-n-1)+b(2an+2a-n^{2}-2n-2)-m]^{2}-{}\\[6pt]&[q+y(a-p-1)+s(2ap+2a-p^{2}-2p-2)-x]^{2}-{}\\[6pt]&[z+p\ell (a-p)+t(2ap-p^{2}-1)-pm]^{2})\\[6pt]\end{aligned}}

Does this polynomial generate a prime number if a=1, b=2, c=3, …, z=26? --114.38.87.55 (talk) 07:55, 30 January 2025 (UTC)[reply]

This polynomial is found in the article. The answer to the question is no. It only produces a nonnegative value if all 14 Diophantine equations are satisfied. As you can read above, with the given value assignment, it fails already on the first equation. --Lambiam 11:14, 30 January 2025 (UTC)[reply]

So what number does this polynomial generate if a=1, b=2, c=3, …, z=26? 111.252.80.160 (talk) 11:36, 30 January 2025 (UTC)[reply]

It'll produce a negative number even though k+2 is 13. All those squares in the big second term need to be zero otherwise it produces zero or a negative number instead of 13 x 1. NadVolum (talk) 12:13, 30 January 2025 (UTC)[reply]

January 31

Why does this algorithm always lead to the trivial square root of $y$ when $y$ is a perfect square?

This is based on Talk:Kunerth's algorithm#Taking The Square Root of 67*Y Mod RSA260
First take $digitsConstant$ , a random semiprime… then use the following pseudocode :

2. Compute : $bb=([[digitsConstant^{0.5}]]+1)^{2}-digitsConstant$
3. Find integers $x$ and $y$ such as (25²+ x×digitsConstant)÷(y×67) = digitsConstant+bb
4. take $z$ , an unknown variable, then expand ((67z + 25)²+ x×digitsConstant)÷(y×67) and then take the last Integer part without a $z$ called $w$ . $w$ will always be a perfect square.
5. $w=sqrt(w)$
6. Find $a$ and $b$ such as a== w (25 + w×b)
7. Solve 0=a²×x²+(2a×b−(x×digitsConstant))×z+(b²−67×y)
8. For each of the 2 possible integer solution, compute z mod digitsConstant.

The fact the result will be a modular square root is expected, but then why if the $y$ computed at step 2 is a perfect square, z mod\ digitsConstant will always be the same as the Integer square root of $y$ and not the other possible modular square ? (that is, the trivial solution). 2A01:E0A:401:A7C0:9CB:33F3:E8EB:8A5D (talk) 09:22, 31 January 2025 (UTC)[reply]

Numbers m such that there is n such that eulerphi(eulerphi(n)) = m but there is no number with exactly m primitive roots

56 is in the range of eulerphi(eulerphi(n)), but there is no number with exactly 56 primitive roots, the numbers like 56 seems to be rare, what is the set of such numbers (i.e. the intersection of (sequence A378508 in the OEIS) and (sequence A231773 in the OEIS)) <= 10000? 220.132.216.52 (talk) 17:16, 31 January 2025 (UTC)[reply]

February 3