## Le Monde puzzle [#959]

**A**nother of those arithmetic Le Monde mathematical puzzle:

Find an integer A such that A is the sum of the squares of its four smallest dividers (including1) and an integer B such that Bis the sum of the third poser of its four smallest factors. Are there such integers for higher powers?

**T**his begs for a brute force resolution checking the integers until a solution appears. The only exciting part is providing the four smallest factors but a search on Stack overflow led to an existing R function:

FUN <- function(x) { x <- as.integer(x) div <- seq_len(abs(x)) return(div[x %% div == 0L]) }

(which uses the 0L representation I was unaware of) and hence my R code:

quest1<-function(n=2){ I=4 stop=TRUE while ((stop)&(I<1e6)){ I=I+1 dive=FUN(I) if (length(dive)>3) stop=(I!=sum(sort(dive)[1:4]^n)) } return(I) }

But this code only seems to work for n=2 as produces A=130: it does not return any solution for the next value of n… As shown by the picture below, which solely exhibits a solution for n=2,5, A=17864 (in the second case), there is no solution less than 10⁶ for n=3,4,6,..9. So, unless I missed a point in the question, the solutions for n>2 are larger if they at all exist.

A resolution got published yesterday night in Le Monde and (i) there is indeed *no* solution for n=3 (!), (ii) there *are* solutions for n=4 (1,419,874) and n=5 (1,015,690), which are larger than the 10⁶ bound I used in the R code, (iii) there is supposedly *no* solution for n=5!, when the R code found that 17,864=1⁵+2⁵+4⁵+7⁵… It is far from the first time the solution is wrong or incomplete!

April 24, 2016 at 10:27 am

[…] Son unas entradas muy interesantes, porque siempre aprendo una función o un truco nuevos. En la entrada del pasado miércoles 20 de abril, Xian atacaba el problema 959, del fin de […]