# Are there more than two rational solutions to a certain system $abcd=a+b+c+d=K$ ($K$ a given constant)?

This question is a follow-on of a previous question asked some days ago which has been deleted due to its lack of precision. In fact, I found it well explained here

But, implicitly, the domain of solutions was numbers with at most two decimals.

Let me restate the question with rational numbers :

Find four rational numbers (it is the domain I am interested in) $$a \le b \le c \le d$$ whose product is equal to their sum, this common value being $$7.11$$ :

$$abcd=a+b+c+d=\frac{711}{100}$$

The solution provided, converted into fractional expressions, is :

$$\begin{cases}a&=&6/5&=&1.20\\b&=&5/4&=&1.25\\c&=&3/2&=&1.50\\d&=&79/25&=&3.16\end{cases}$$

The numbers involved aren't so surprizing because of the prime factor decompositions :

$$711=3^2 \times 79 \ \ \text{and} \ \ 100=2^2 \times 5^2.$$

Using a computer "screening" and the above decompositions, I have found another solution :

$$\begin{cases}a&=&81/100&=&0.81\\b&=&5/3&=&1.666...\\c&=&2&&\\d&=&79/30&=&2.6333...\end{cases}$$

My questions : are there other (rational) solutions and how can we find them ? Beyond that, is there some theory behind (like the theory of rational points on cubic curves) ?

• Surprisingly, asking on the web "four numbers whose sum is equal to their product 7.11", I realized that this question has been asked a certain number of times under different forms. I am currently examining them. Commented Jun 3 at 7:24
• Commented Jun 3 at 7:29
• Commented Jun 3 at 7:34
• @Gerry Myerson Thanks ... but they treat the case of rational solutions in general but not with the value $7.11$. Commented Jun 3 at 8:10
• @Jean Marie,Okay, give me some time. Commented Jun 4 at 7:01

$$abcd=a+b+c+d=K$$

Substitute $$d=K-a-b-c$$ into $$abcd=K,$$ we get $$-abc^2+ab(K-a-b)c-K=0$$ For the quadratic in $$c$$ to have rational solutions, the discriminant must be a rational square. Hence, this problem is reduced to the problem of finding the rational solutions for the quartic equation. $$v^2 = a^4b^2+b(2b^2-2bK)a^3+b(bK^2+b^3-2b^2K)a^2-4abK$$ b and K are given rational numbers.

               (a,b,c,d), K=711/100

158/205, 75/41, 41/20, 123/50
237/200, 6/5, 8/5, 25/8
252/325, 7/4, 395/182, 169/70
243/325, 79/39, 25/12, 169/75
15/19, 237/133, 49/25, 361/140
16/15, 121/100, 81/44, 395/132
395/448, 48/35, 147/64, 64/25
21/23, 161/100, 23/14, 474/161
75/98, 474/245, 49/25, 49/20
1185/1219, 6/5, 2809/1150, 529/212
22/25, 11/8, 632/275, 225/88
15/14, 79/63, 243/140, 686/225
77/100, 21/11, 150/77, 869/350
1343/1600, 25/17, 153/64, 1024/425
207/275, 1975/1012, 23/11, 1331/575
1369/1400, 6/5, 600/259, 3871/1480
8/9, 158/117, 169/72, 6561/2600
2844/2465, 841/680, 1156/725, 25/8
539/600, 11/7, 400/231, 6399/2200
28/33, 2765/1914, 841/350, 9801/4060
20/19, 20449/16150, 85557/48620, 867/286
85557/100100, 10/7, 24843/10450, 605/247