## Le Monde puzzle [#1063] A simple (summertime?!) arithmetic Le Monde mathematical puzzle

1. A “powerful integer” is such that all its prime divisors are at least with multiplicity 2. Are there two powerful integers in a row, i.e. such that both n and n+1 are powerful?
2.  Are there odd integers n such that n² – 1 is a powerful integer ?

The first question can be solved by brute force.  Here is a R code that leads to the solution:

```isperfz <- function(n){
divz=primeFactors(n)
facz=unique(divz)
ordz=rep(0,length(facz))
for (i in 1:length(facz))
ordz[i]=sum(divz==facz[i])
return(min(ordz)>1)}

lesperf=NULL
for (t in 4:1e5)
if (isperfz(t)) lesperf=c(lesperf,t)
twinz=lesperf[diff(lesperf)==1]
```

with solutions 8, 288, 675, 9800, 12167.

The second puzzle means rerunning the code only on integers n²-1…

``` 8
 288
 675
 9800
 235224
 332928
 1825200
 11309768
```

except that I cannot exceed n²=10⁸. (The Le Monde puzzles will now stop for a month, just like about everything in France!, and then a new challenge will take place. Stay tuned.)

### 6 Responses to “Le Monde puzzle [#1063]”

1. Courtheyn Joel Says:

In the solution of the second part, 675 should not be in the list as it is 26^2-1 and 26 is an even number.

• xi'an Says:

Thanks, I forgot to add the odd constraint!

2. […] article was first published on R – Xi’an’s Og, and kindly contributed to […]

3. FOULLEY Jean-Louis Says:

Problem probably inspired by the Math exam this year for the Bachelor degree Section S , Math Speciality , Exercise No 4

• xi'an Says:

4. Seems the first question would be answered by the second one? As $n^2$ is always powerful, then $n^2 - 1$ and $n^2$ are pair integers that satisfy the first question.
And we can show there are infinitely many: for any $n^2 - 1$ which is a powerful integer, $(2n^2 - 1)^2 - 1 = 4n^2(n^2 - 1)$ will also be a powerful integer.