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22:01
@Ante Can you prove it for special choices of $a$ and $b$ for *ANY $f(x)$ (that assumes a value of zero at least once) ?
22:13
i have found a way for general a and b but it is too tiresome to write all of that here
i can highlight the idea
hello @Peter
Please highlight it
well chose that f is linear on [a,c] with f(a)=0, and for example f(c)=-1, then choose f to be linear on [c,b] so that $\int_{a}^{b}{f(x)dx}=f(b)$

you have equation to determine c then
if that works
if not f(c) need to be chosen differently
although that could not work if f(c) is not chosen appropriately
f should be made of two linear functions
if f linear does not work then try with parabola
it should work for appropriately chosen f(c)
or, better, define f(b)=b and determine c and f(c)
you have f(a)=0 and f(b)=b and $\int_{a}^{b}{f(x)dx}=b$
so c and f(c) need to be determined
I have to understand your work, it’s tough :-)
22:33
@Ante Hi, still no result. Intuitively, I would now guess that there are large squarefree solutions.
why
the prime factors of sigma(n) and phi(n) are different from the prime factors of n, and I see nothing that prevents the sum being divisible by n.
sigma(n)/n is also unbounded
Isn't it reasonable to consider the chance for a solution to be 1/n ? Like in the case of Wieferich primes ?
Can we construct any equation not too weird with sigma(n) and phi(n) that is not already in OEIS ?
It makes more fun to analyze something new.
n*phi(n)=sigma(n)?
I noticed phi(n) | sigma(n)
but I do not know whether this is in OEIS
2phi(n) + 3sigma(n) = 6n ?
nah, (n+phi(n))|sigma(n) is better
22:44
? for(n=1,10^7,if(2*eulerphi(n)+3*sigma(n)==6*n,print1(n," ")))
14 442 1377
? for(n=1,10^7,if(Mod(sigma(n),n+eulerphi(n))==0,print1(n," ")))
2 456 828 7584 33462 1357440 1596048 1964544
?
are those in OEIS ?
the last one is not
Great ! And the first is ?
nope, also not
OK, yours is in fact more interesting.
yes
22:47
First, I will collect some solutions and then begin with conditions.
for (n+phi(n))|sigma(n) ?
yes
Solutions so far :
? for(n=1,10^8,if(Mod(sigma(n),n+eulerphi(n))==0,print(n," ",sigma(n)/(n+eulerphi(n))," ",factor(n))))
2 1 Mat([2, 1])
456 2 [2, 3; 3, 1; 19, 1]
828 2 [2, 2; 3, 2; 23, 1]
7584 2 [2, 5; 3, 1; 79, 1]
33462 2 [2, 1; 3, 2; 11, 1; 13, 2]
1357440 3 [2, 7; 3, 1; 5, 1; 7, 1; 101, 1]
1596048 2 [2, 4; 3, 1; 41, 1; 811, 1]
1964544 2 [2, 9; 3, 1; 1279, 1]
19800384 2 [2, 6; 3, 1; 281, 1; 367, 1]
26211264 2 [2, 6; 3, 1; 211, 1; 647, 1]
31451136 2 [2, 11; 3, 1; 5119, 1]
largest fraction at the moment : 3
Are all solutions even ?
complete upto 10^8
106805184 2 [2, 6; 3, 1; 167, 1; 3331, 1]
OK, good night. I will think abouth this !
bye
23:25
@Knight I have already provided you a proof about yesterday at Mathematics chat
About that yesterday*
@palindromicprime Hi!
23:43
something new to research?
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