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Consider the Diophantine equation $x^3+x+y^3+y = z^3 + z$ for positive integer $x,y,z$.

I tried small values and got some near equalities : $(5,6,7)$ and $(12,16,18)$ are true up to value $2$. $( 5^3 + 5 + 6^3 + 6 = 7^3 + 7 + 2 )$.

$(6,8,9)$ is true up to value $4$. And it appears that there are infinite near misses having a value of $80$ or $12$ or a power of $2$.

I already tried to find a contradiction mod $30$ and mod $7$ but did not find any.

I noticed that $x^3 + x$ can be squarefree and I cannot find infinite descent.

Does the polynomial ($x^3+x+y^3+y -(z^3 + z)$) factor perhaps ? It is irreducible in $Z$ though.

I must note that I can solve the similar looking $x^3 -x + y^3 -y = z^3 - z$ over the integers which is also irreducible over $Z$. (for instance by letting $z = u+v$ and $x = u - v$ reducing a subset of the solutions to a Pell equation )

Maybe Im close. I would like to solve this Diophantine without using p-adic.

And I also wonder about $x^3 + x + y^3 + y - z^3 - z = (12,80,2^n)$ and $x^5 + x + y^5 + y = z^5 + z$ for $x,y,z>0$.

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1 Answer 1

up vote 4 down vote accepted

Some 'brute force' solutions of $\ x^3 + x + y^3 + y - z^3 - z = r$
(with additional contributions from Oleg567)

For $\ r=0\ $ (and $z<10^5$) : \begin{array} {cc|c} x&y&z\\ \hline 36& 37& 46\\ 98& 248& 253\\ 165& 705& 708\\ 320& 377& 442\\ 843& 1078& 1228\\ 2372& 3323& 3685\\ 2988& 3070& 3817\\ 6963& 8320 & 9703 \\ 8997 & 20636 & 21191 \\ 24338 & 27272 & 32617 \\ 28783 & 29763 & 36892 \\ 39697 & 50635 & 57728 \\ 41488 & 66006 & 71071 \\ 55682 & 74983 & 84072 \\ \end{array}

(for general solutions see $16$. of part 11 of Tito Piezas III 'Collection of Algebraic Identities')

For $\ r=12$ : \begin{array} {cc|c} x&y&z\\ \hline 1& 2& 0\\ 72& 141& 147\\ \end{array}

For $\ r=80$ : \begin{array} {cc|c} x&y&z\\ \hline 11& 20& 21\\ 18& 30 &32 \\ 216& 644& 652\\ 704& 781& 938\\ \end{array}

For $\ r=2^0\ $ none found ($\max(x,y)<10^4$).

For $\ r=2^1$ : \begin{array} {cc|c} x&y&z\\ \hline 1& u& u\\ 5& 6& 7\\ 12& 16& 18\\ 101& 218& 225\\ 215& 377& 399\\ \end{array}

For $\ r=2^2$ : \begin{array} {cc|c} x&y&z\\ \hline 1& 1& 0\\ 6& 8& 9\\ \end{array}

For $\ r=2^3$ : \begin{array} {cc|c} x&y&z\\ \hline 9& 10& 12\\ 17& 40& 41\\ 92& 509& 510\\ 568& 1104& 1152\\ 2177& 2599& 3032\\ \end{array}

For $\ r=2^4$ : \begin{array} {cc|c} x&y&z\\ \hline 9& 15& 16\\ 130& 199& 216\\ 214& 274& 312\\ \end{array}

For $\ r=2^5$ : \begin{array} {cc|c} x&y&z\\ \hline 1573& 2897& 3044\\ 4857& 7281& 7940\\ \end{array}


I found no solution to $\ x^5 + x + y^5 + y = z^5 + z$
(in the reduced range checked $\max(x,y)<10^4$).

Similar kinds of equations were much studied at the beginning of the 20th century. Dickson's famous books "History Of The Theory Of Numbers" may be consulted online (and downloaded) at archive.org. Perhaps that the methods proposed around these pages could inspire you... or Tito Piezas III since he appears from time to time here or at MO.

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LOL i checked upto x = 35 ! However I think you missed solutions for r=12 and r=80. Thanks for the answer. However I would like a 'theory' too. Brute force is the computer beating the human. ( I did not look at Tito's page at the time of comment ) There is also r = 2^n left. –  mick Nov 17 '12 at 18:50
    
@Mick: yeah but I was cautious (I wrote 'some') and for $2^n$ I'll add some answers too (don't hope an infinity... ;-)). –  Raymond Manzoni Nov 17 '12 at 20:38
    
Hehe I do not hope for infinity in a finite world. –  mick Nov 17 '12 at 20:53
1  
ok. You (please). and I will delete my comment after. I tried to find solution for 5-th degree, but without success too. –  Oleg567 Nov 19 '12 at 23:25
    
@Oleg567: Perhaps that we may prove that none exist at all for the 5-th degree (after all the OP wants formulas and proofs...). Thanks for the addition anyway, –  Raymond Manzoni Nov 19 '12 at 23:32

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