Properties of $F(p)=p^2+1$, where $p$ is a prime number

Question

Let $F(p)=p^2+1$, where $p$ is a prime number. For what primes $p_1$, $p_2$ does $p_2$ divide $F(p_1)$ and $p_1$ divide $F(p_2)$? Two examples are $\{p_1,p_2\}=\{5,13\}$, and $\{p_1,p_2\} = \{89,233\}$. Are there other pairs that have this property?

Answer 1

ADDED: Eric Towers found an additional two pair of primes. Numbered as my computer output below, they are

===========================

218:  529892711006095621792039556787784670197112759029534506620905162834769955134424689676262369 P 
219:  1387277127804783827114186103186246392258450358171783690079918032136025225954602593712568353 P
287:  36684474316080978061473613646275630451100586901195229815270242868417768061193560857904335017879540515228143777781065869 P 
288:  96041200618922553823942883360924865026104917411877067816822264789029014378308478864192589084185254331637646183008074629 P 

++======================

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.