For which positive integers $\ j\ge 3\ $ do the numbers $\ a:=\varphi(j)\ $ and $\ b:=\frac{\varphi(j+1)}{2}\ $ where $\ \varphi(n)\ $ is the totient-function, give a solution of a primitive pythagorean triple, that is we have $\ \gcd(a,b)=1\ $ and $\ a^2+b^2=c^2\ $ for some positive integer $\ c\ $ ?
I found the following solutions upto $\ j=10^8\ $
? for(j=2,10^8,a=eulerphi(j);b=eulerphi(j+1)/2;if(gcd(a,b)==1,if(issquare(a^2+b^2)==1,print([j,a,b]))))
[8, 4, 3]
[21, 12, 5]
[30, 8, 15]
[24178, 9360, 12089]
?
$\ a\ $ is even, hence $\ b\ $ must be odd, so $\ j+1\ $ cannot have more than one odd prime factor. Neither can $\ j+1\ $ be divisible by $\ 4\ $ and have an odd prime factor. Finally, $\ j+1\ $ cannot be divisible by $\ 8\ $. Is the list of solutions I found complete ?