**A** number theory Le Monde mathematical puzzle whose R coding is not really worth it *(and which rings a bell of a similar puzzle in the past, puzzle I cannot trace…)*:

*The set Ξ is made of pairs of integers (x,y) such that (i) both x and y are written as a sum of two squared integers (i.e., are bisquare numbers) and (ii) both xy and (x+y) are bisquare numbers. Why is the product condition superfluous? For which values of (a,b) is the pair (13*^{a},13^{b}) in *Ξ *?

**I**n the first question, the property follows from the fact that the product of two bisquare numbers is again a bisquare number, thank to the remarkable identity

(a²+b²)(c²+d²) = (ac+bd)²+(ad-bc)²

(since the double products cancel). For the second question, once I realised that

13=2²+3²

it followed that any number 13^{a} was the sum of two squares, hence a bisquare number, and thus that the only remaining constraint was that (b≥a)

13^{a+}13^{b}=13^{a}(1+13^{b-a})

is also bisquare. If b-a is *even*, this sum is then the product of two bisquare numbers and hence a bisquare number. If b-a is *odd*, I do not have a general argument to bar the case (it certainly does not work for 13+13² and the four next ones).

Filed under: Books, Kids, R Tagged: Le Monde, mathematical puzzle, number theory

*Related*

To

**leave a comment** for the author, please follow the link and comment on their blog:

** Xi'an's Og » R**.

R-bloggers.com offers

**daily e-mail updates** about

R news and

tutorials on topics such as:

Data science,

Big Data, R jobs, visualization (

ggplot2,

Boxplots,

maps,

animation), programming (

RStudio,

Sweave,

LaTeX,

SQL,

Eclipse,

git,

hadoop,

Web Scraping) statistics (

regression,

PCA,

time series,

trading) and more...

If you got this far, why not

__subscribe for updates__ from the site? Choose your flavor:

e-mail,

twitter,

RSS, or

facebook...