I have no complete answer this week, question 3 was beyond my
Questions 1 and 2.
For convenience, rearrange the given equality as floor(m*s/n) =
I claim that it holds for all positive integers s<n<m iff m-s <=
essentially improving the bound m-s < sqrt(m) by a constant factor
If m-s <= 1, there are no integers n with s<n<m; let us assume
m-s >= 2.
From the definition of floor function, we obtain the following
system of inequalities:
(a) m*s/n >= m-n+s-1;
(b) m*s/n < m-n+s.
Multiply by n (strictly positive) and rearrange:
(a) n^2 - (m+s-1)*n + m*s >= 0;
(b) n^2 - (m+s)*n + m*s < 0;
Let us call the quadratic polynomials Pa(n) and Pb(n) respectively.
Pb(n) has two distinct integer roots s and m, so inequality (b)
holds for s<n<m, precisely the range of interest.
Therefore, the starting equality holds in such range iff inequality
(a) holds too.
Pa(n) has discriminant Da = (m+s-1)^2 - 4*m*s;
I will prove that (a) holds for all integers s<n<m (actually, for
all positive integers n) iff Da <= 1.
We need to distinguish two cases.
I) Da is even, m+s is odd, m-s >= 3.
Pa(n) has a unique minimum at the integer point n1 = (m+s-1)/2, with
Pa(n1) = -Da/4 and s<n1<m;
iff Da<=0, inequality (a) holds for n1, and for any other positive
II) Da is odd, m+s is even, m-s >= 2.
Now n1 is not an integer; Pa(n) has two minima at the closest
integer points n0 = (m+s)/2 and n2 = (m+s-2)/2 = n0-1,
with Pa(n0) = Pa(n2) = (1-Da)/4 and s<n0<m (but s<n2<m holds too
only if m-s>=4);
iff Da<=1, inequality (a) holds for n0, and for any other positive
So, inequality Da <= 1 actually covers both parity cases.
Rewrite it as s^2 - 2*(m+1)*s + m*(m-2) <= 0, and solve for s:
m+1-sqrt(4*m+1) <= s <= m+1+sqrt(4*m+1),
or, after rearranging:
-sqrt(4*m+1)-1 <= m-s <= sqrt(4*m+1)-1.
But only the second inequality is meaningful, as we worked in the
range m-s >= 2.
As done before, we can rewrite the given equality as a system of two
(a) p(n+2)*p(n)/p(n+1) > p(n+2)-p(n+1)+p(n)-1;
(b) p(n+2)*p(n)/p(n+1) < p(n+2)-p(n+1)+p(n).
Now both inequalities are strict, because p(n+1) doesn't divide the
product p(n+2)*p(n) exactly.
Again, (b) holds iff p(n)<p(n+1)<p(n+2), a true statement for
Multiply both sides of (a) by p(n+1), then rearrange:
(p(n+2)-p(n+1))*(p(n+1)-p(n)) < p(n+1).
Assuming the conjectured Oppermann's bound on prime gaps:
p(n+1)-p(n) < sqrt(p(n)) for every n > 30,
we obtain (for n> 30) the following chain of inequalities:
(p(n+2)-p(n+1))*(p(n+1)-p(n)) < sqrt(p(n+1))*sqrt(p(n)) < p(n+1).
It is easy to check that, in the range n <= 30, (a) fails only for
n=3 and for n=8.
The given inequality can be obtained from previous inequality (a),
by solving for p(n+2).
Take two positive integers s, m such that s < m
Consider the function f(x) = m - x + s - m*s / x for x
between s and m (s < x < m).
With a littlebit of elementary calculus we can show that f(x)
reaches its maximum where f '(x) = 0, i.e. : when x0 = Sqrt(m*s).
We then have :
f(x0) = m - Sqrt(m*s) + s - Sqrt(m*s) = m + s -
2Sqrt(m)Sqrt(s) = (Sqrt(m) - Sqrt(s))^2.
Now, if we want to have m - n + s -Floor[m*s / n] to be equal
to 1 for all integers n between s and m
we should have that m - x + s - m*s / x is a number N that
satisfies 0 < N <= 1.
Thus, that should be Sqrt(m) - Sqrt(s) <= 1.
This is much more than m - Sqrt(m) < s.
For the rest, the condition Sqrt(p(n+2)) - Sqrt(p(n) < 1 would
imply that there is always a prime in an interval [ x^2,
As far as I know this is still a conjecture.
So, I think all subsequent
inequalities are also conjectural.
For all n, with s<n<m and s,n,m integer we are given: m-n+s-floor[m*s/n]=1
Write: Floor[m*s/n]=m*s/n-a with a in [0,1)
We thus obtain: m-n+s-m*s/n=1-a=b, with b in (0,1]
Multiply with n and simplify: (m-n)(n-s)/n=b
The function f(x,s,m)=(m-x)(x-s)/x is 0 iff x=m or x=s. Its maximum
value is attained for x=sqrt(ms).
Given s<n<m we won’t have x=s or x=m, thus b>0. We want this maximum
value b to be less than or equal to 1 for all values of n.
We therefore need:
(m-sqrt(ms))(sqrt(ms)-s)/ sqrt(ms) = (sqrt(m)-sqrt(s))^2<=1 or
sqrt(m)-sqrt(s)<=1 (since m>s).
If we multiply both sides of the last equation with sqrt(m)+sqrt(s)
we arrive at:
The reverse of this statement is that if we have m-s<=sqrt(m)+sqrt(s)
m-n+s-Floor[m*s/n]=1 for all n with: s<n<m.
The proof just reverses every step and is left as an exercise.
That means that the given condition in Q1: m-s<sqrt(m) is too
strict. It is sufficient, but not necessary.
The equation: m-n+s-Floor[m*s/n]=1 with s<n<m may have particular
solutions n even if m-s>sqrt(m)+sqrt(s), but it can’t have solutions
for all n in the given interval.
This might happen if m-n or n-s is relatively small.
It is very tempting to plug in: s=p[n], n=p[n+1], m=p[n+2]. But we
can’t use our previous result. In order to be able to do that we
definitely need to prove that we have:
p[n+2]-p[n] < sqrt(p[n+2])+sqrt(p[n])
But I don’t see how to prove that this condition applies. It might
for large n, but even that is just a conjecture nowadays: Opperman’s
conjecture, which is a statement regarding an average prime gap, for
p[n+1]-p[n] < sqrt(p[n])
Add the two ‘inequalities’ and arrive at:
So it seems to fit nicely, but there are two drawbacks. One
Opperman’s is a conjecture and not a theorem. And two the conjecture
is about an average prime gap and not a prime gap. This does not
mean that the given equality we wish to use is not true; as stated
before it might also happen that m-n or n-s are small enough, or to
put it differently, at least one of the two prime gaps needs to be
If we plug in the same values for s,n,m into (m-n)(n-s)/n<=b with
b=1 and rewrite we arrive at:
Note the possibility of equality above. And again, this statement
might very well be true, with a few exceptions, but I don’t see how
to prove the condition stated in my answer of Q3.
He hecho la demostración del
es equivalente a probar:
0<(n-s)(m-n) los cual es obvio
por otro lado
la cota correcta sería (m-s)<Sqrt[n]
en lugar de Sqrt[m]
por lo que se tendría (m-n)(n-s)<(m-s)^2<=n
y así estarían probadas la dos
Para el recíproco buscamos un
contraejemplo con (m-s)>Sqrt[n]
no he encontrado ninguno poniendo
como cota Sqrt[n]+1, tampoco para Sqrt[n]+2, he
encontrado contraejemplos para Sqrt+3
m=5 n=2 s=1
por lo que la mejor cota sería
pero esto último no lo puedo