Show that if $a,b,c\in \mathbb{N}$ and ${a^2+b^2+c^2}\over{abc+1}$ is an integer it is the sum of two nonzero squares

I was reading about the fascinating problem in the IMO ($$1988$$ #$$6$$) that asks:

Let $$a$$ and $$b$$ be positive integers such that $$(ab+1) | (a^2+b^2)$$. Show that $${a^2+b^2}\over{ab+1}$$ is a perfect square.

I wondered what would happen with the natural extension of this problem to three constants. Upon exploration with Mathematica, I found that the integral values of $${a^2+b^2+c^2}\over{abc+1}$$ all seemed to be expressible as the sum of two nonzero perfect squares. The converse also seems to be true (given that a number is the sum of two nonzero squares, it seems to be expressible in the form of this fraction).

So I pose two questions:

If $$a,b,c\in \mathbb{N}$$ and $$k={{a^2+b^2+c^2}\over{abc+1}}$$ is an integer, is $$k$$ the sum of two nonzero squares?

and

Given that $$k$$ is expressible as the sum of two nonzero perfect squares, do there exist $$a,b,c\in\mathbb{N}$$ for which $${{a^2+b^2+c^2}\over{abc+1}}=k$$?

• Does the usual method for the $(a^2+b^2)(ab+1)$ problem also work here? – Lord Shark the Unknown Jan 31 at 2:48
• I unfortunately don't have enough experience to be able to solve that – volcanrb Jan 31 at 2:50
• These days, this problem can be googled as well as solved: Searching for "IMO Problem 6" leads to en.wikipedia.org/wiki/Vieta_jumping – Lord Shark the Unknown Jan 31 at 2:59
• To whoever downvoted: how can I improve the question? – volcanrb Jan 31 at 3:11

I PROVED THE CONJECTURE

$$x^2 + y^2 + z^2 = k + kxyz$$ Notice that we cannot have a solution with $$x<0$$

Got the other part. Let $$x \geq y \geq z \geq 1$$ be a solution giving $$k$$ that minimizes $$x+y+z$$ among positive solutions giving $$k.$$ Vieta jumping is just the new solution $$(x,y,z) \mapsto (kyz-x,y,z)$$ The equation is $$x^2 - (kyz)x + y^2 + z^2 - k = 0.$$ The replacement root $$x'$$ gives $$x + x' = kyz.$$ The product is $$xx' = y^2 + z^2 - k.$$

We have minimized $$x+y+z$$ among positive triples giving $$k.$$ We get a new triple with $$x' = kyz - x.$$ As it is impossible to have $$x' <0,$$ we have an Alternative:

(A) $$kyz - x = 0$$

(B) $$kyz - x \geq x$$

We will show that alternative (B) does not occur for, say, $$k > 2.$$

ASSUME (B), namely $$kyz \geq 2x,$$ with $$x \geq y \geq z \geq 1.$$ $$x^2 + y^2 + z^2 = k + (kyz)x \geq k + (2x)x= k + 2x^2$$ $$\color{red}{ y^2 + z^2 \geq k + x^2}.$$ But $$y^2 - (kx)yz + z^2 = k - x^2.$$ Add, $$2y^2 - (kx)yz + 2z^2 \geq 2k .$$ $$y^2 - \left(\frac{kx}{2} \right)yz + z^2 \geq k .$$ So, $$y \geq z \geq 1$$ and $$\color{red}{y^2 - \left(\frac{kx}{2} \right)yz + z^2 > 0 }.$$ From the quadratic formula, $$y > \frac{1}{4} \left( kx + \sqrt {k^2 x^2 - 16} \right) z .$$ When $$k \geq 4,$$ we see $$y > x,$$ a contradiction. When $$k=3$$ and $$z \geq 2,$$ we still get $$y > x.$$ Finally, with $$k=3$$ and $$z=1,$$ the condition $$y^2 + z^2 \geq k + x^2$$ reads $$y^2 + 1 \geq 3 + x^2$$ or $$y^2 \geq 2 + x^2,$$ so again $$y > x.$$

These contradictions tell us that alternative (A) actually holds for $$k \geq 3.$$ That is, $$kyz = x,$$ so $$kyz - x = 0,$$ and we have a new solution $$(0,y,z)$$ with $$y \geq z \geq 1$$ and this $$x=0$$ to $$x^2 + y^2 + z^2 = k + kxyz.$$ As this $$x=0,$$ we have reached $$\color{red}{ y^2 + z^2 = k }$$

The method is called Vieta Jumping. It is a special case of the automorphism group of an indefinite binary quadratic form.

Part Two

The first part of your question: given $$k = a^2 + b^2,$$ you get an answer with $$x=a, \; \; y =b, \; \; z = kab$$ Then $$\frac{x^2 + y^2 + z^2}{1+xyz} = k$$

Part One

The description of Vieta Jumping in links is not quite specific enough to settle this. The conjecture seems very likely; here are the values of $$k$$ with $$5000 \geq a \geq b \geq c \geq 1,$$ but ignoring the many, many ways to get $$k=2.$$

     k         a         b         c         k
5        10         2         1         5 =  5
5      1102       230         1         5 =  5
5       230        48         1         5 =  5
5      2399        48        10         5 =  5
5        48        10         1         5 =  5
5      4948        99        10         5 =  5
5       980        99         2         5 =  5
5        99        10         2         5 =  5
8        32         2         2         8 =  2^3
8       510        32         2         8 =  2^3
10      2940       297         1        10 =  2 5
10       297        30         1        10 =  2 5
10        30         3         1        10 =  2 5
10       899        30         3        10 =  2 5
13      2025        78         2        13 =  13
13      3040        78         3        13 =  13
13        78         3         2        13 =  13
17      1152        68         1        17 =  17
17      4623        68         4        17 =  17
17        68         4         1        17 =  17
18       162         3         3        18 =  2 3^2
20       160         4         2        20 =  2^2 5
25       300         4         3        25 =  5^2
26       130         5         1        26 =  2 13
26      3375       130         1        26 =  2 13
29       290         5         2        29 =  29
32       512         4         4        32 =  2^5
34       510         5         3        34 =  2 17
37       222         6         1        37 =  37
40       480         6         2        40 =  2^3 5
41       820         5         4        41 =  41
45       810         6         3        45 =  3^2 5
50      1250         5         5        50 =  2 5^2
50       350         7         1        50 =  2 5^2
52      1248         6         4        52 =  2^2 13
53       742         7         2        53 =  53
58      1218         7         3        58 =  2 29
61      1830         6         5        61 =  61
65      1820         7         4        65 =  5 13
65       520         8         1        65 =  5 13
68      1088         8         2        68 =  2^2 17
72      2592         6         6        72 =  2^3 3^2
73      1752         8         3        73 =  73
74      2590         7         5        74 =  2 37
80      2560         8         4        80 =  2^4 5
82       738         9         1        82 =  2 41
85      1530         9         2        85 =  5 17
85      3570         7         6        85 =  5 17
89      3560         8         5        89 =  89
90      2430         9         3        90 =  2 3^2 5
97      3492         9         4        97 =  97
98      4802         7         7        98 =  2 7^2
100      4800         8         6       100 =  2^2 5^2
101      1010        10         1       101 =  101
104      2080        10         2       104 =  2^3 13
106      4770         9         5       106 =  2 53
109      3270        10         3       109 =  109
116      4640        10         4       116 =  2^2 29
122      1342        11         1       122 =  2 61
125      2750        11         2       125 =  5^3
130      4290        11         3       130 =  2 5 13
145      1740        12         1       145 =  5 29
148      3552        12         2       148 =  2^2 37
170      2210        13         1       170 =  2 5 17
173      4498        13         2       173 =  173
197      2758        14         1       197 =  197
226      3390        15         1       226 =  2 113
257      4112        16         1       257 =  257
290      4930        17         1       290 =  2 5 29
k         a          b         c        k