Properties of $F(p)=p^2+1$, where $p$ is a prime number
ADDED: Eric Towers found an additional two pair of primes. Numbered as my computer output below, they are
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218: 529892711006095621792039556787784670197112759029534506620905162834769955134424689676262369 P
219: 1387277127804783827114186103186246392258450358171783690079918032136025225954602593712568353 P
287: 36684474316080978061473613646275630451100586901195229815270242868417768061193560857904335017879540515228143777781065869 P
288: 96041200618922553823942883360924865026104917411877067816822264789029014378308478864192589084185254331637646183008074629 P
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no opinion on prime $x,y.$
We have integers $(1+x^2)/ y$ and $(1+y^2)/x.$ We see that both $x,y$ divide $1 + x^2 + y^2.$ In turn, this says that $\gcd(x,y)=1;$ if a prime $p | x$ and $p | y,$ then $p|1$ which is a contradiction.
We have reached $xy | 1 + x^2 + y^2 $ in positive integers.
It follows that
$$ 3xy = x^2 + y^2 + 1 $$
This is Problem 2 in Yimin Ge
See also https://en.wikipedia.org/wiki/Vieta_jumping#Constant_descent_Vieta_jumping
So, positive integers with $$ \color{blue}{ x^2 - 3xy + y^2 = -1 } $$
There are concrete ways to find all integer pairs with $x^2 - 3xy + y^2 = -1.$ In particular, with fixed target $-1,$ there is just a single orbit of pairs...
I probably should not have numbered the lines below. Beginning with 1,1, these are $x_{n+2} = 3 x_{n+1} - x_n$
Thu Dec 2 17:21:56 PST 2021
4 2 P
5 5 P
6 13 P
7 34
8 89 P
9 233 P
10 610
11 1597 P
12 4181
13 10946
14 28657 P
15 75025
16 196418
17 514229 P
18 1346269
19 3524578
20 9227465
21 24157817
22 63245986
23 165580141
24 433494437 P
25 1134903170
26 2971215073 P
27 7778742049
28 20365011074
29 53316291173
30 139583862445
31 365435296162
32 956722026041
33 2504730781961
34 6557470319842
35 17167680177565
36 44945570212853
37 117669030460994
38 308061521170129
39 806515533049393
40 2111485077978050
41 5527939700884757
42 14472334024676221
43 37889062373143906
44 99194853094755497 P
45 259695496911122585
46 679891637638612258
47 1779979416004714189
48 4660046610375530309
49 12200160415121876738
50 31940434634990099905
51 83621143489848422977
52 218922995834555169026
53 573147844013817084101
54 1500520536206896083277
55 3928413764606871165730
56 10284720757613717413913
57 26925748508234281076009
58 70492524767089125814114
59 184551825793033096366333
60 483162952612010163284885
61 1264937032042997393488322
62 3311648143516982017180081
63 8670007398507948658051921
64 22698374052006863956975682
65 59425114757512643212875125
66 155576970220531065681649693
67 407305795904080553832073954
68 1066340417491710595814572169 P
69 2791715456571051233611642553
70 7308805952221443105020355490
71 19134702400093278081449423917 P
72 50095301248058391139327916261
73 131151201344081895336534324866
74 343358302784187294870275058337
75 898923707008479989274290850145
76 2353412818241252672952597492098
77 6161314747715278029583501626149
78 16130531424904581415797907386349
79 42230279526998466217810220532898
80 110560307156090817237632754212345
81 289450641941273985495088042104137
82 757791618667731139247631372100066
83 1983924214061919432247806074196061
84 5193981023518027157495786850488117
85 13598018856492162040239554477268290
86 35600075545958458963222876581316753
87 93202207781383214849429075266681969
88 244006547798191185585064349218729154
89 638817435613190341905763972389505493
90 1672445759041379840132227567949787325
91 4378519841510949178490918731459856482
92 11463113765491467695340528626429782121
93 30010821454963453907530667147829489881
94 78569350599398894027251472817058687522
95 205697230343233228174223751303346572685
96 538522340430300790495419781092981030533
97 1409869790947669143312035591975596518914
98 3691087032412706639440686994833808526209
99 9663391306290450775010025392525829059713
100 25299086886458645685589389182743678652930
101 66233869353085486281758142155705206899077
102 173402521172797813159685037284371942044301
103 453973694165307953197296969697410619233826
Next Day. Here is my proof that $\frac{1+x^2 + y^2}{xy}$ must equal $3.$
LEMMA
Given integers $$ m > 0, \; \; M > m+2, $$ there are no integers $x,y$ with $$ x^2 - Mxy + y^2 = -m. $$
PROOF
Calculus: $m+2 > \sqrt{4m+4},$ since $(m+2)^2 = m^2 + 4m + 4,$ while $\left( \sqrt{4m+4} \right)^2 = 4m + 4.$ Therefore also $$ M > \sqrt{4m+4} $$
We cannot have $xy < 0,$ as then $x^2 - M xy + y^2 \geq 2 + M > 0. $ It is also impossible to have $x=0$ or $y=0.$ From now on we take integers $x,y > 0.$
With $x^2 - Mxy + y^2 < 0,$ we get $0 < x^2 < Mxy - y^2 = y(Mx - y),$ so that $Mx - y > 0$ and $y < Mx.$ We also get $x < My.$
The point on the hyperbola $ x^2 - Mxy + y^2 = -m $ has both coordinates $x=y=t$ with $(2-M) t^2 = -m,$ $(M-2)t^2 = m,$ and $$ t^2 = \frac{m}{M-2}. $$ We demanded $M > m+2$ so $M-2 > m,$ therefore $t < 1.$ More important than first appears, that this point is inside the unit square.
We now begin to use the viewpoint of Hurwitz (1907). All elementary, but probably not familiar. We are going to find integer solutions that minimize $x+y.$ If $2 y > M x,$ then $y > Mx-y.$ Therefore, when Vieta jumping, the new solution given by $$ (x,y) \mapsto (Mx - y, x) $$ gives a smaller $x+y$ value. Or, if $2x > My,$ $$ (x,y) \mapsto (y, My - x) $$ gives a smaller $x+y$ value. We already established that we are guaranteed $My-x, Mx-y > 0.$
Therefore, if there are any integer solutions, the minimum of $x+y$ occurs under the Hurwitz conditions for a fundamental solution (Grundlosung), namely $$ 2y \leq Mx \; \; \; \; \mbox{AND} \; \; \; \; 2 x \leq My. $$ We now just fiddle with calculus type stuff, that along the hyperbola arc bounded by the Hurwitz inequalities, either $x < 1$ or $y < 1,$ so that there cannot be any integer lattice points along the arc. We have already shown that the middle point of the arc lies at $(t,t)$ with $t < 1.$ We just need to confirm that the boundary points also have either small $x$ or small $y.$ Given $y = Mx/2,$ with $$ x^2 - Mxy + y^2 = -m $$ becomes $$ x^2 - \frac{M^2}{2} x^2 + \frac{M^2}{4} x^2 = -m, $$ $$ x^2 \left( 1 - \frac{M^2}{4} \right) = -m $$ $$ x^2 = \frac{-m}{1 - \frac{M^2}{4}} = \frac{m}{ \frac{M^2}{4} - 1} = \frac{4m}{M^2 - 4}. $$ We already confirmed that $ M > \sqrt{4m+4}, $ so $M^2 > 4m+4$ and $M^2 - 4 > 4m.$ As a result, $ \frac{4m}{M^2 - 4} < 1.$ The intersection of the hyperbola with the Hurwitz boundary line $2y = Mx$ gives a point with $x < 1.$ Between this and the arc middle point, we always have $x < 1,$ so no integer points. Between the arc middle point and the other boundary point, we always have $y < 1.$ All together, there are no integer points in the bounded arc. There are no Hurwitz fundamental solutions. Therefore, there are no integer solutions at all.