For what kind of positive integers $m+n+m^2+n^2 \mid m^3+n^3$ holds?
Solution 1:
I'll summarise progress here to prevent the comments becoming hard to read. I have made this community wiki, so feel free to add any progress made.
Parametric solutions found (by Ivan Neretin): $$\begin{align}m = k(2k+1), &\quad n=k(2k-1) \\ m = k(k-1)(k^2+k-1), &\quad n=k(k-1)(k+1)\end{align}$$ Solutions found outside of the above patterns (covers all solutions for $n \leq m \leq 2300000$):
m n T/B d a b
30 6 28 6 5 1
1002 654 897 6 167 109
2052 864 1872 108 19 8
2145 1023 1936 33 65 31
3192 1218 2940 42 76 29
3258 1926 2912 18 181 107
3388 1188 3146 44 77 27
5768 2296 5292 56 103 41
6273 451 6242 41 153 11
6372 918 6260 54 118 17
10872 6408 9720 72 151 89
16302 4636 15428 38 429 122
18208 12512 16380 32 569 391
20193 13515 18126 159 127 85
21414 6142 20252 166 129 37
33462 5742 32668 198 169 29
34086 6118 33212 874 39 7
34476 6708 33462 156 221 43
39015 25785 34992 135 289 191
44520 11928 42336 168 265 71
44940 25620 40200 420 107 61
55870 12580 53780 370 151 34
81054 53082 72657 54 1501 983
87363 6273 86946 153 571 41
92736 37408 84992 224 414 167
105490 63910 94325 770 137 83
111120 50160 100800 240 463 209
111948 79572 101080 228 491 349
123820 44280 114800 820 151 54
150048 57312 138240 288 521 199
168280 113960 151200 280 601 407
173652 35148 168200 348 499 101
182505 45195 174570 345 529 131
193830 110370 173394 390 497 283
200784 84372 183312 1068 188 79
218880 112896 196608 2304 95 49
268362 106182 246408 306 877 347
274533 205275 249696 357 769 575
290475 39525 285912 75 3873 527
306360 63940 296240 460 666 139
360780 193620 323400 420 859 461
372372 65100 363258 2604 143 25
405790 186186 367598 4774 85 39
415443 113953 394346 511 813 223
445018 85652 432180 1862 239 46
503244 256212 452392 36 13979 7117
520740 143550 494100 990 526 145
599280 465168 548856 528 1135 881
719448 101928 707296 744 967 137
733800 389400 658125 600 1223 649
802740 237660 757200 1020 787 233
911820 211380 876096 780 1169 271
1005420 304980 946400 780 1289 391
1043610 620490 933100 1290 809 481
1070895 790305 971960 705 1519 1121
1195947 955413 1102240 747 1601 1279
1299892 560604 1183952 1364 953 411
1359402 1203312 1290828 2838 479 424
1420860 243780 1387200 1020 1393 239
1536210 99990 1530150 9090 169 11
1543302 688734 1401372 1494 1033 461
1776373 521883 1676696 379 4687 1377
1777440 219360 1754064 480 3703 457
1835856 673200 1698048 1584 1159 425
1969620 845580 1794690 1020 1931 829
2034960 283560 2001600 16680 122 17
2171580 1255620 1942080 1020 2129 1231
2222580 1816620 2059992 1020 2179 1781
2271720 326040 2232450 1320 1721 247
Necessary conditions (by Geoff Robinson): Either $m$ and $n$ share a common factor, or they are both odd. If they are both odd, then $m+n \equiv 0$ mod $4$.
Moreover, $m$ and $n$ have the same parity. Proof: The denominator $B=m^2+n^2+m+n$ is even. The numerator $T=m^3+n^3$ is a multiple of $B$, so is even, so $m^3$ and $n^3$ have the same parity, so $m$ and $n$ have the same parity.
This means that if $d$ is odd, $a$ and $b$ are odd. I (Rosie F) notice that in most cases $d$ is even. But it is quite rare that either $a$ or $b$ is even.
We also can remark that the only coprime solution is $(m,n)=(3,1)$. To show this assume $(m,n)=1$. Also note that $m^3+n^3=(m+n)(m^2+n^2-mn)$, so the original quantity is an integer iff $$m^2+n^2+m+n\mid (m+n)(m+n+mn)$$ Of course this is true iff $m^2+n^2+m+n\mid \gcd(m^2+n^2+m+n, m+n)(m+n+mn)$, which simplifies to $m^2+n^2+m+n\mid \gcd(m+n, 2mn)(m+n+mn)$. Much like Geoff noted, this can only be the case when $\gcd(m+n, 2mn)>1$ since $m^2+n^2+m+n>m+n+mn$. However under the assumption that $\gcd(m,n)=1$, this is only the case when $m$ and $n$ are both odd and $\gcd(m+n, 2mn)=2$. Hence our original quantity is an integer only if $$m^2+n^2+m+n\mid 2(m+n+nm)$$ and $m$ and $n$ are odd. Assume $m^2+n^2+m+n\neq 2(m+n+nm)$, then for some prime $p$, $m^2+n^2+m+n\mid 2(m+n+nm)/p$ and thus $$m^2+n^2+m+n\le 2(m+n+nm)/p\le m+n+nm$$ which is false. Hence $m^2+n^2+m+n=2(m+n+nm)$ and $(m-n)^2=n+m$. This gives $m-n\mid n+m$ so $$m-n=\gcd(m-n,n+m)=\gcd(m+n, 2m)=2$$ Thus $m=n+2$. Subbing this into $(m-n)^2=m+n$ gives $(m,n)=(3,1)$. Thus if $\gcd(m,n)=1$ we must have $(m,n)=(3,1)$, which if we test works.
Solution 2:
For the equation.
$$x^3+y^3=q(x^2+y^2+x+y)$$
We use the solutions of the equation Pell.
$$p^2-(3k-2t)(2t-k)s^2=1$$
Then the solution can write.
$$x=((3k-2t)s+p)2ts$$
$$y=\frac{t}{k-t}(p^2+2(2k-t)ps+(3k-2t)ks^2)$$
$$q=\frac{t^2}{k(k-t)}(p^2+2(3k-2t)ps+(3k-2t)^2s^2)$$
And again.
$$y=((3k-2t)s-p)2ts$$
$$q=\frac{4(3k-2t)}{k}t^2s^2$$
All the numbers can have any sign.