Archive for congruence

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Diophantine riddle

Posted in Books, Kids, R with tags , , , , , , , on February 27, 2023 by xi'an

The weekly riddle from The Riddler is to find solutions to the Diophantine equation

c³-c=b²+4

(when b and c are positive integers). First, forget about ChatGPT since it states this is a Pell equation. With a wrong argument. Second, when running a basic R code, using as.double to handle larger integers, the only solution less than 10⁶ this code returned was

[1]    999799 999700015

with the first column being c and the second b. But this is not a correct solution!, as confirmed by Mathematica, which states there is no integer solution. This makes sense when looking at the unique real solution (in c) of the cubic

c³-c-(b²+4)=0

since the solution (using Cardano’s formula) involves

\sqrt[3]{\frac{b^2+4}{2}\pm\sqrt{\frac{(b^2+4)^2}{4}-\frac{1}{27}}}

leaving the inverse of 27 as the only non-integer term in the expression when b is even… (The exact solution that this Diophantine equation has no solution is simpler: the lhs is a multiple of 3, while the rhs cannot be, as shown by looking at b(3).)

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shortened iterations [code golf]

Posted in Kids, pictures, Statistics, Travel with tags , , , , , , , , on October 29, 2019 by xi'an

A codegolf lazy morning exercise towards finding the sequence of integers that starts with an arbitrary value n and gets updated by blocks of four as

a_{4k+1} = a_{4k} \cdot(4k+1)\\ a_{4k+2} = a_{4k+1} + (4k+2)\\ a_{4k+3} = a_{4k+2} - (4k+3)\\ a_{4k+4} = a_{4k+3} / (4k+4)

until the last term is not an integer. While the update can be easily implemented with the appropriate stopping rule, a simple congruence analysis shows that, depending on n, the sequence is 4, 8 or 12 values long when

n\not\equiv 1(4)\\ n\equiv 1(4)\ \text{and}\ 3(n-1)+4\not\equiv 0(32)\\ 3(n-1)+4\equiv 0(32)

respectively. But sadly the more interesting fixed length solution

`~`=rep #redefine function
b=(scan()-1)*c(32~4,8,40~4,1,9~3)/32+c(1,1,3,0~3,6,-c(8,1,9,-71,17)/8)
b[!b%%1] #keep integers only

ends up being longer than the more basic one:

a=scan()
while(!a[T]%%1)a=c(a,d<-a[T]*T,d+T+1,e<-d-1,e/((T<-T+4)-1))
a[-T]

where Robin’s suggestion of using T rather than length is very cool as T has double meaning, first TRUE (and 1) then the length of a…

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an attempt at code golf

Posted in Kids, R with tags , , , , , , , , , on May 15, 2019 by xi'an

Having discovered codegolf on Stack Exchange a few weeks ago, I spotted a few interesting puzzles since then but only got the opportunity at a try over a quiet and rainy weekend (and Robin being on vacation)! The challenge was to write an R code for deciding whether or not a given integer n is congruent or not, when congruent means that it is the surface of a rectangle triangle with all three sides rational. The question included a pointer to the Birch and Swinnerton-Dyer conjecture as a mean to check congruence although the real solution was provided by Tunnell’s Theorem, which states that n is congruent if and only if the number of integer solutions to 2x²+y²+8z²=n is twice as much as the number of integer solutions to 2x²+y²+32z²=n if n is odd and  the number of integer solutions to 8x²+y²+16z²=n is twice as much as the number of integer solutions to 8x²+y²+64z²=n if n is even. Although this is only true for squared-free integers. (I actually spent more time on figuring out the exact wording of the theorem than on optimising the R code!)

My original solution

p=function(n){
 for (i in(n:2)^2)if(n%%i<1)n=n/i
 if(n%%2){d=8;f=2;g=16}else{d=2;f=1;g=8}
 A=0;b=(-n:n)^2
 for(x in d*b)for(y in x+f*b)for(z in g*b)
  A=A+(y+z==n)-2*(y+4*z==n)
 A==0}

was quite naïve, as shown by the subsequent improvements by senior players, like the final (?) version of Guiseppe:

function(n){b=(-n:n)^2
for(i in b[b>0])n=n/i^(!n%%i)
P=2^(n%%2)
o=outer
!sum(!o(y<-o(8/P*b,2*b,"+")/P-n,z<-16/P*b,"+"),-2*!o(y,4*z,"+"))}

exhibiting a load of code golf tricks, from using an anonymous function to renaming functions with a single letter, to switching from integers to booleans and back with the exclamation mark.

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Le Monde puzzle [#960]

Posted in Kids, R with tags , , , , , on April 28, 2016 by xi'an

An arithmetic Le Monde mathematical puzzle:

Given an integer k>1, consider the sequence defined by F(1)=1+1 mod k, F²(1)=F(1)+2 mod k, F³(1)=F²(1)+3 mod k, &tc. [With this notation, F is not necessarily a function.] For which value of k is the sequence the entire {0,1,…,k-1} set?

This leads to an easy brute force resolution, for instance writing the R function

[sourcecode language=”r” gutter=”false”]
crkl<-function(k) return(unique(cumsum(1:(2*k))%%k))
[/sourcecode]

where 2k is a sufficient substitute for ∞. Then the cases where the successive images of 1 visit the entire set {0,1,…,k-1} are given by

[sourcecode language=”r” gutter=”false”]
> for (i in 2:550) if (length(crkl(i))==i) print(i)
[1] 2
[1] 4
[1] 8
[1] 16
[1] 32
[1] 64
[1] 128
[1] 256
[1] 512
[/sourcecode]

which suspiciously looks like the result that only the powers of 2 k=2,2²,2³,… lead to a complete exploration of the set {0,1,…,k-1}. Checking a few series in the plane back from Warwick, I quickly found that when k is odd, (1) the sequence is of period k and (2) there is symmetry in the sequence, which means it only takes (k-1)/2 values. For k even, there is a more complicated symmetry, with the sequence being of period 2k, symmetric around its two middle values, and taking the values 1,2,..,1+k(2k+1)/4,..,1+k(k+1)/2. Those values cannot cover the set {0,1,…,k-1} if two are equal, which means an i(i+1)/2 congruent to zero modulo k, hence equal to k. This is clearly impossible when k is a power of 2 because i and i+1 cannot both divide a power of 2. I waited for the published solution as of yesterday’s and the complete argument was to show that when N=2p, the corresponding sequence [for N] is made (modulo p) of the sequence for p plus the same sequence translated by p. The one for N is complete only if the one for p is complete, which by recursion eliminates all cases but the powers of 2…

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Le Monde puzzle [#29]

Posted in R, Statistics with tags , , on July 29, 2011 by xi'an

This week, the puzzle from the weekend edition of Le Monde was easy to state: in the sequence (8+17n), is there a 6th power? a 7th? an 8th? If so, give the first occurrence. So I first wrote an R code for a function testing whether an integer is any power:

[sourcecode language=”r” gutter=”false”]
ispower=function(x){
ispo=FALSE
logx=log(x)
i=trunc(logx/log(2))
while((i>1)&&(!ispo)){
j=t=trunc(exp(logx/i))
while (t<x) t=j*t
ispo=(x==t)
if (!ispo){
j=t=j+1
while (t<x) t=j*t
ispo=(x==t)}
i=i-1}
list(is=ispo,pow=j)}
[/sourcecode]

(The function returns the highest possible power.) Then I ran the thing over the first million of values of the sequence:

[sourcecode language=”r” gutter=”false”]
fib=8
for (j in 1:10^6){
fib=fib+17
tes=ispower(fib)
if (tes$is)
print(c(fib,tes$pow,log(fib)/log(tes$pow)))}
[/sourcecode]

only to find that only the powers 2,3,6,10,11,19 were present among the first terms. Continue reading →

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