It is Problem #4, page 16 mentioned in the cited work. Edited from the manuscript.
The author's[WS] solution is on page 38. (The text is written in green.) In translation:
"ANSWER : All numbers in the arithmetic progression
$65k+56,\;\;k=0,\;1,\;2\dots$
[I noticed at the bottom of the page in the first photo that there is no indication of how to find the answer. Then comes its Solution :]
Indeed, if $n=65k+56$ , $k\geqslant 0$ is an integer, then $n\equiv 1\;(mod\;5)$ and $n\equiv 4\;(mod\;13)$ from where $4n^2+1\equiv 0\;(mod\;5)$ and $4n^2+1\equiv 0\;(mod\;13)$, so that
$5\mid 4n^2+1$ and $13\mid 4n^2+1.$
$\color {Green}{\blacksquare}$"
I solved the problem without knowing this answer. [Text written in blue; in translation:]
ANSWER CiP
The numbers that have the property in the statement are exactly those of the form
$65k+4,\;\;65k+9,\;\;65k+56,\;\;65k+61$ where $k=0,\;1,\;2,\dots\;.$
Solution CiP
Since $5$ and $13$ are coprime, we have
$5\mid A\;\;and\;\;13 \mid A\;\;\;\Leftrightarrow\;\;\;5\cdot 13 \mid A$
Let $n=65k+r$ ; then $4n^2+1=4(65k+r)^2+1=4(65^2k^2+2\cdot 65\cdot r+r^2)+1=$
$=65\cdot(4\cdot 65k^2+8r)+4r^2+1=65k_1+4r^2+1.$
Trying to choose a number $r$ so that $4r^2+1=65$ we get $r^2=16$. For example $r=4$, so we have an infinity of numbers, of the form $n=65k+4$, with the property in the statement. The problem would be solved [, but not completely.]
$\color {Blue}{\blacksquare}$
I noted in red, further on, that we can also have $r=-4$, obtaining another infinity of convenient numbers.
- And all the convenient values found in my answer were obtained by checking all the possibilities
$65k,\;65k\pm1,\;65k\pm2,\;\dots,\;65k\pm32$
Finally, I also noted, [written in black], that : The numbers in ANSWER CiP form the sequence
A203464 in OEIS ("The On-Line Encyclopedia of Integer Sequences).
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