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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.

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  • $\begingroup$ Interesting problem. Where is it from? $\endgroup$ – user21820 Jul 20 '14 at 7:05
  • $\begingroup$ Note that dividing, by $abcd$, it's $${1\over a}+{1\over b}+{1\over c}+{1\over d}={1\over abcd}$$ $\endgroup$ – Gerry Myerson Jul 20 '14 at 8:51
  • $\begingroup$ This problem came as a bi-product of trying to understand the first few pages of Saveliev's "Invariants of Homology Spheres." @ Gerry Myerson Is that a hint? $\endgroup$ – thedude Jul 20 '14 at 10:31
  • $\begingroup$ Just an observation that might, or might not, be useful. By the way, @ Gerry doesn't work --- it has to be @Gerry to notify me. $\endgroup$ – Gerry Myerson Jul 20 '14 at 11:10
  • $\begingroup$ On the solution of the equation: $abc+abd+acd+bcd=k$ What is known? $\endgroup$ – individ Jul 23 '14 at 10:47
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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.

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An infinite family of integer solutions:

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

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  • 1
    $\begingroup$ Wow, how did you get this??? $\endgroup$ – Gina Jul 20 '14 at 7:20
  • $\begingroup$ i am sure this question has circulated in few books in number theory or some "problem solving" journals... $\endgroup$ – DeepSea Jul 20 '14 at 7:33
  • $\begingroup$ @8pi r If you can find those references, that would be great. $\endgroup$ – thedude Jul 20 '14 at 10:31
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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.

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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.

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

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

$$c=-ts$$

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

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