Does anybody have any ideas how to solve the equation $x = \varphi(x) + \varphi(x + 1) – 1$, where $x$ is a natural number and $\varphi$ is Euler's totient function?
I failed even to figure out whether this equation has finite number of solutions or infinite.
Any help will be appreciated.
On
Thius is more a comment than anything else.
As said in comments, using your numbers, I found that this is sequence $A067798$ at $OEIS$.
Continuing the search up to $10^7$, I got the following numbers $$\{1,2,3,5,9,15,21,35,39,45,75,99,135,231,255,363,483,765,855,1295,1599,2015,2115,43 35,6783,9999,14399,16095,16599,18495,30495,53823,62799,63455,65535,77615,155319, 186999,196095,327675,589815,686735,722015,732735,1456719,1679615,2559999,4064255 \}$$
Some more observations, not an answer but too large for a comment.
If $x$ is a prime, then:
$$\varphi(x) + \varphi(x + 1) - 1 = x$$ $$x - 1 + \varphi(x + 1) - 1 = x$$ $$\varphi(x + 1) = 2$$
So we can not have prime $x > 5$. If $x$ is $n^2 - 1$, with even $n$ then:
$$\varphi(x) + \varphi(x + 1) - 1 = x$$ $$\varphi(n^2 - 1) + \varphi(n^2) - 1 = n^2 - 1$$ $$\varphi((n+1)(n-1)) + n\varphi(n) = n^2$$ $$\varphi(n+1)\varphi(n-1) + n\varphi(n) = n^2$$ $$\varphi(n+1)\varphi(n-1) = n(n - \varphi(n))$$
Using this equality I found the following (non-exhaustive) set of numbers:
$$\{3, 15, 35, 99, 255, 483, 1295, 1599, 2115, 9999, 14399, 18495, 53823, 65535, 732735, 1679615, 2559999, 4064255, 18800895, 207359999, 259081215, 930006015, 3943839999, 4294967295, 6024243455, 471606333695\}$$
The last $n$ for which the above equality holds below $10^7$ is $686736$.
Using a similar technique for $n^4-1$ I've found $x = 353473690402815$ as the biggest solution yet.