Diophantine equation abc + abd + acd + bcd= 1

Question

Is there a reference which classifies or at least gives an infinite family of integer solutions to the above equation? A solution to the problem would also be great obviously.

Answer 1

An infinite family of integer solutions:

$a = -n;\\ b = n + 1; \\c = n^2 + n + 1;\\ d = (n^2 + n)^2 + n^2 + n + 1.$

Answer 2

Equation:

$$abc+abd+acd+bcd=1$$

Has another solution.

$$a=1-k$$

$$b=k^2-k+1$$

$$c=k$$

$$d=k^4-2k^3+2k^2-k+1$$

$$...............$$

$$a=1-k$$

$$b=k^2-k-1$$

$$c=k$$

$$d=-k^4+2k^3-k-1$$

$$...............$$

$$a=-(k+1)$$

$$b=-(k^2+k+1)$$

$$c=k$$

$$d=-k^4-2k^3-2k^2-k+1$$

$$...............$$

$$a=-(k+1)$$

$$b=-(k^2+k-1)$$

$$c=k$$

$$d=k^4+2k^3-k-1$$

$k$ - integer of any sign.

Answer 3

This is D28 in Guy, Unsolved Problems In Number Theory. He writes,

Mordell asked for the integer solutions of $${1\over w}+{1\over x}+{1\over y}+{1\over z}+{1\over wxyz}=0$$ Several papers have appeared, giving parametric families of solutions. For example, Takahiro Nagashima sends solutions ... $(1366,-15,7,-13)$ ... and more generally $w=xyz+1$ with $$x=-2eh^3-deh^2(n-3)+eh(n-1-2de)+1,\\y=2deh^2+eh(n-3)-de(n-1)+1,\\z=-2deh^2-eh(n-1)-1$$ where $d,e=\pm1$ independently, but there seems to be no guarantee that these four two-parameter families give all solutions.

Guy also gives a number of references, of which the most recent would be Chan Wah-Keung, Solutions of a Mordell Diophantine equation, J Ramanujan Math Soc 6 (1991) 129-140, MR 93d:11033.

There is also Choudhry, Ajai, A Diophantine equation of Mordell, J. Ramanujan Math. Soc. 24 (2009), no. 2, 113–126, MR2543546 (2010e:11026). The review by Lajos Hajdu says, "In the present paper the author gives infinitely many two-parameter solutions of the equation, which are more general than the parametric solutions previously found. Further, it is also shown how even more general solutions can be obtained."

The Mordell reference, by the way, is L. J. Mordell, Canad. Math. Bull., 17 (1974) 149.

Answer 4

I think it is better to rewrite the solution for the more General case.

$$abc+abd+acd+bcd=q$$

The decision will write it down.

$$a=k+t$$

$$b=\frac{k^2+tk+p}{t}$$

$$d=-k$$

$$c=\frac{(k(k+2t)+p+t^2)k^2+tpk+tq}{tp}$$

$k$ - integer of any sign. $t,p$ - select no fraction.

It is possible differently. Choose any number "$p$" and the expression lay at the multipliers.

$$p^2+1=ks$$

Then the solution can be written.

$$a=-p$$

$$b=p+s$$

$$c=p+k$$

$$d=p(p+s)(p+k)+q$$

If you choose "$t,k$" - the numbers so that they were not Pythagorean triples.

We use the solutions of the equation Pell. $p^2-(t^2+k^2)s^2=1$

Decisions will write it down.

$$a=p+(t-k)s$$

$$b=p+(t+k)s$$

$$c=-ts$$

$$d=ts(p+(t+k)s)(p+(t-k)s)+q$$