I was randomly browsing, when I found this puff piece claiming a proof of the Lindelöf hypothesis by Fokas. Note that the Wikipedia article says that he claimed, then withdrew his claim in 2017, but the USC piece is dated June 25 2018. So, what is the truth?
[Math] Lindelöf hypothesis claim
analytic-number-theoryca.classical-analysis-and-odesnt.number-theory
Related Solutions
I don't know the best way to express RH inside PA, but the following inequality $$\sum_{d \mid n} d \leq H_n + \exp(H_n)\log(H_n),$$ where $H_n = 1+1/2+\cdots+1/n$ is the $n$-th harmonic number, is known to be equivalent to RH. [J. Lagarias, An elementary problem equivalent to the Riemann hypothesis, Amer. Math. Monthly, 109 (2002), 5347–543.] The same paper mentions another inequality of Robin, $$\sum_{d \mid n} d \leq e^\gamma n \log\log n \qquad (n \geq 5041),$$ where $e^\gamma = 1.7810724\ldots$, which is also equivalent to RH. Despite the appearance of $\exp,$ $\log$ and $e^\gamma$, it is a routine matter to express these inequalities as $\Pi^0_1$ statement. (Indeed, the details in Lagarias's paper make this even simpler than one would originally think.)
We address the problem of Euler, showing an asymptotic lower bound for the number of ways of writing $n\equiv 2 \pmod 4$ as a sum of four squares $a^2+b^2+c^2+d^2$ where neither $a^2+b^2$ nor $c^2+d^2$ is divisible by any prime $\equiv 3 \pmod 4$. For simplicity we shall only count the solutions where $a^2+b^2\equiv c^2+d^2 \equiv 1\pmod 4$.
Let $r(n)$ denote the number of ways of writing $n$ as a sum of two squares so that $r(n) = 4 \sum_{d|n} \chi_{-4}(d)$ with $\chi_{-4}$ being the non-trivial character $\pmod 4$. Clearly the number of solutions we want is at least $$ \sum_{\substack{{a+b=n} \\ {a\equiv b\equiv 1 \pmod 4}}} r(a)r(b) - 2 \sum_{p\equiv 3 \pmod 4} \sum_{\substack{{a+b=n} \\ {a\equiv b\equiv 1\pmod 4} \\ {p^2 |a} }} r(a) r(b) = S-2T, $$ say. (The first sum counts all possible solutions with $a\equiv b\equiv 1 \pmod 4$ and the second sum excludes all solutions with either $a$ or $b$ being divisible by the square of sum prime $\equiv 3 \pmod 4$, and by symmetry we may assume that $a$ is the one divisible by $p^2$.)
Let us focus just on evaluating the sum $T$; the sum $S$ is simpler and may be handled similarly. Write
the sum $T$ as $\sum_{p \equiv 3 \pmod 4} T(p)$, so that (since $r(ap^2)=r(a)$)
$$
T(p) = \sum_{\substack{ {a p^2 +b =n} \\ {a\equiv b\equiv 1\pmod 4}}} r(a) r(b).
$$
Since $r(k)\ll k^{\epsilon}$ for all $k$, clearly $T(p) \ll n^{1+\epsilon}/p^2$, so that large primes $p$
contribute a negligible amount. Now let us suppose that $p$ is small.
Now if $a\equiv 1 \pmod 4$ we have
$$
r(a) = 4\sum_{k\ell =a} \chi_{-4}(a) = 8\sum_{\substack{ {k|a} \\ {k\le \sqrt{a} } }}^{\prime} \chi_{-4}(k),
$$
using that $\chi_{-4}(k)=\chi_{-4}(\ell)$ (because $k\ell\equiv 1\pmod 4$), and the prime
over the sum indicates that the term $k=\sqrt{a}$ is counted with weight $1/2$.
Therefore
$$
T(p) =8 \sum_{k}^{\prime} \chi_{-4}(k) \sum_{\substack{{b=n-p^2 k \ell} \\ {b\equiv 1\pmod 4} \\ {\ell \ge k}} } r(b).
$$
Now the sum over $b$ above is just a sum over values $b$ that are at most $n-p^2 k^2$ and
satisfying the congruence conditions $b\equiv 1\pmod 4$ and $b\equiv n \pmod{p^2k}$.
In other words, this is a problem of understanding the distribution of the sums of two squares
function in arithmetic progressions.
Sums of two squares in arithmetic progressions. Let us quickly recall what is known about this problem. Let $R(x;q,a)$ be the sum of $r(n)$ over all $n\le x$ with $n\equiv a \pmod q$. The key result we need is that there is a good asymptotic formula for $R(x;q,a)$ uniformly for $q$ up to $x^{2/3-\epsilon}$. This is due to Selberg and Hooley (unpublished), with details first appearing in a paper of R. A. Smith. For a recent nice version see Tolev's paper in Proc. Steklov Inst. (2012, vol 276, 261--274), whose version we shall use. Let $\eta_a(q)$ denote the number of solutions to the congruence $x^2+y^2 \equiv a \pmod q$ with $1\le x, y \le q$. Then $$ \sum_{\substack{ {n\le x} \\ {n\equiv a\pmod q}} } r(n) = \pi \frac{\eta_a(q)}{q^2} x + O((q^{\frac 12} + x^{\frac 13}) (a,q)^{\frac 12} x^{\epsilon}). $$ Note that this bound uses the Weil bound for Kloosterman sums; but for our application we need only a bound that permits $q$ a little larger than $\sqrt{x}$, and less precise elementary estimates for Kloosterman sums would suffice for this.
On the function $\eta_a(q)$. Note also that
$\eta_a(q)$ is a multiplicative function of $q$, and is always $\ll q d(q)$ (where $d(q)$ is the number
of divisors of $q$). In fact it is not hard to compute what $\eta_a(q)$ is, and as we shall need this below we summarize this calculation. If $p$ is an odd prime not dividing $a$ then we may show that
$\eta_a(p^{k}) = p^{k-1} (p-\chi_{-4}(p))$. If $p$ is an odd prime that does divide $a$, then
to compute $\eta_a(p^k)$ we classify the solutions to $x^2+y^2 \equiv a \pmod {p^k}$ as non-singular
if $p$ doesn't divide one of $x$ or $y$ and singular otherwise. The non-singular solutions are
seen to number $2p^{k-1}(p-1)$ if $p\equiv 1\pmod 4$ and $0$ if $p\equiv 3 \pmod 4$. As for the singular solutions, if $k=1$ then this number is just $1$; if $k\ge 2$ and $p^2 \nmid a$ there
are no singular solutions; and if $k\ge 2$ and $p^2|a$ then the number of singular solutions is
$p^2 \eta_{a/p^2}(p^{k-2})$. This describes fully how to compute $\eta_a(q)$. From this description and
some calculation we find that
$$
\sum_{k=1}^{\infty} \chi_{-4}(k) \frac{\eta_a(k)}{k^{1+s}} = \frac{L(s,\chi_{-4})}{\zeta_2(s+1)} {\tilde \sigma}_s(n)
$$
where $\zeta_2(s) = \zeta(s) (1-2^{-s})$ (the Euler factor at $2$ removed), and
$$
{\tilde \sigma}_s(n) = \sum_{\substack{{d|n}\\ {d \text{ odd}} } } d^{-s}.
$$
Given a prime $p\equiv 3 \pmod 4$, for our work on $T(p)$ we shall also need to understand
the following related Dirichlet series:
$$
\sum_{k=1}^{\infty} \chi_{-4}(k) \frac{\eta_a(kp^2)}{k^{1+s}}
$$
and by a similar calculation we see that if $p\nmid a$ this equals
$$
p(p+1)\Big(1-\frac{1}{p^{s+1}}\Big)^{-1} \frac{L(s,\chi_{-4})}{\zeta_2(s+1)} {\tilde \sigma}_s(n);
$$
and if $p|a$ but $p^2\nmid a$ it is zero; and if $p^2|a$ it equals
$$
p^2 \frac{L(s,\chi_{-4})}{\zeta_2(s+1)} {\tilde \sigma}_s(n/p^2) .
$$
Back to $T(p)$. We now use the asymptotic formula above in our expression for $T(p)$. Since $\eta_1(4) = 8$ we obtain that $$ T(p) = 4\pi \sum_{k \le \sqrt{n}/p} \chi_{-4}(k) \Big( \frac{\eta_n(p^2k)}{p^4 k^2} (n-p^2k^2) + O\Big((p\sqrt{k} +n^{\frac 13}) (n,p^2k)^{\frac 12} n^{\epsilon}\Big)\Big). $$ There are three cases: either $p\nmid n$, $p|n$ but $p^2\nmid n$, and $p^2|n$.
Consider the first case when $p\nmid n$. Here the error terms may be bounded easily as $$ O\Big( \frac{n^{\frac 56+\epsilon}}{p} + \frac{n^{\frac 34+\epsilon}}{\sqrt{p}}\Big). $$ Now consider the main term which is $$ 4\pi \frac{n}{p^4} \sum_{k\le \sqrt{n}/p} \chi_{-4}(k) \frac{\eta_n(p^2k)}{k^2} \Big(1 - \frac{p^2k^2}{n}\Big). $$ We can express this as a contour integral (for some suitably large $c$) $$ 4\pi \frac{n}{p^4} \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Big(\sum_{k=1}^{\infty} \chi_{-4}(k) \frac{\eta_n(p^2k)}{k^{2+s}}\Big) \Big( \frac{\sqrt{n}}{p}\Big)^s \frac{2}{s(s+2)}ds. $$ Using our knowledge of the Dirichlet series above, we get that the above is $$ 4\pi \frac{n}{p^4} \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} p (p+1) \Big(1-\frac{1}{p^{2+s}}\Big)^{-1} \frac{L(1+s,\chi_{-4})}{\zeta_2(s+2)} {\tilde \sigma}_{1+s}(n) \Big(\frac{\sqrt{n}}{p}\Big)^s \frac{2}{s(s+2)} ds. $$ Now moving the line of integration to Re$(s)=-1+\epsilon$ we obtain that the above is $$ 4\pi \frac{n}{p^4} \Big( p(p+1) \Big(1-\frac{1}{p^2}\Big)^{-1} \frac{L(1,\chi_{-4})}{\zeta_2(2)} {\tilde \sigma}_1(n) + O\Big( p^3 n^{-\frac 12+\epsilon}\Big) \Big). $$ Since $L(1,\chi_{-4}) =\pi/4$ and $\zeta_2(2) = \pi^2/8$ the above is $$ 8 \frac{n}{p^2} \Big(1-\frac{1}{p}\Big)^{-1} {\tilde \sigma}_1(n) + O\Big( \frac{n^{\frac 12+\epsilon}}{p}\Big). $$ Thus in this case we find that $$ T(p) = 8\frac{n}{p^2} \Big(1-\frac{1}{p}\Big)^{-1} {\tilde \sigma}_1(n) + O\Big( \frac{n^{\frac 56+\epsilon}}{p} + \frac{n^{\frac 34+\epsilon}}{\sqrt{p}} \Big). $$ Recall also that $T(p) \ll n^{1+\epsilon}/p^2$ which is useful if $p$ is very large.
In the second case when $p|n$ but $p^2\nmid n$, trivially $T(p)=0$.
Finally if $p^2|n$ then we carry out a similar argument to the one above. This gives $$ T(p) = 8\frac{n}{p^2} {\tilde \sigma}_1(n/p^2) + O \Big(n^{\frac 56+\epsilon} +\sqrt{p} n^{\frac 34+\epsilon} \Big). $$ Once again $T(p)\ll n^{1+\epsilon}/p^2$ which is useful for large $p$.
Putting all our work above, and splitting into the cases $p\le n^{1/6}$ (where we use our asymptotic formula) and $p>n^{1/6}$ (where we use the trivial bound $\ll n^{1+\epsilon}/p^2$) we obtain that $$ T(p) = 8n {\tilde \sigma}_1(n) \sum_{\substack{{p\equiv 3 \pmod 4} \\ {p\nmid n}}} \frac{1}{p(p-1)} + 8n \sum_{\substack{{p\equiv 3 \pmod 4} \\ {p^2 |n}}} \frac{1}{p^2} {\tilde \sigma}_1(n/p^2) +O(n^{\frac 56+\epsilon}). $$
Conclusion. In the same way, we find that $$ S = 8n {\tilde \sigma}_1(n) + O(n^{\frac 56+\epsilon}). $$ It follows that the number of solutions to Euler's question is at least $$ S-2T \ge 8n {\tilde \sigma}_1(n) \Big( 1- \sum_{p\equiv 3 \pmod 4} \frac{2}{p(p-1)}\Big) + O(n^{\frac 56+\epsilon}) \ge 4n {\tilde \sigma}_1(n) + O(n^{\frac 56+\epsilon}). $$
Remarks. With more work one can establish an asymptotic in Euler's problem rather than just the
lower bound. Similarly one can give an asymptotic in Hermite's problem. In principle one can make
all the error terms explicit, and possibly obtain a complete result for all $n$, but this would entail a lot of work.
Lastly in connection with Hermite's problem, a classical problem of Hardy and Littlewood asks for the representations of numbers as $p+x^2+y^2$ with $p$ a prime. This was resolved by Hooley on GRH, and Linnik unconditionally. Once the Bombieri-Vinogradov theorem became available, the result became much simpler.
This could be modified to give a stronger version of Hermite's problem for large $n$, by also asking for $p$ to be
$1\pmod 4$ -- a cross between Hermite and Goldbach.
Best Answer
(Not an answer of any sort, just too long for a comment.) The main result seems to be an integral equation (1.3) of the form $$\int_{-\infty}^\infty K(t,\tau) |\zeta(\tfrac{1}{2}+it\tau)|^2\,d\tau={\mathcal G}(t)$$ with some explicit functions $K$ and ${\mathcal G}$. This equation (if true) is presumably new, and may be interesting.
However, in my view, how interesting it is would depend quite a bit on whether $|\zeta(\tfrac{1}{2}+it\tau)|^2$ is the only solution of it. This sort of integral operators may have kernels, and if it is the case then it would be rather difficult to squeeze the Lindelöf hypothesis out of it. (If I were the author then this is where I would look.)
P.S. For those who have read ``puff piece'': As pointed out by Robert Israel, indeed, in the (this far, latest) version 4 a proof of the Lindelöf hypothesis is not claimed.
P.P.S Take the above with a pinch of salt; the last time I was involved with this subject was decades ago.