When does $a^2+b^2 = c^3 +d^3$ for all integer values $(a, b, c, d) \ge 0$. I believe this only happens when: $a^2 = c^3 = e^6$ and $b^2 = d^3 = f^6$. With the following exception:
Would that statement be correct?
Is there a general formula for when this happens?
On
Not really natural to demand both cubes positive...
1 1 2 = 2
1 0 1 = 1
2 2 16 = 2^4
2 1 9 = 3^2
2 0 8 = 2^3
3 -1 26 = 2 * 13
4 4 128 = 2^7
4 2 72 = 2^3 * 3^2
4 1 65 = 5 * 13
4 0 64 = 2^6
4 -3 37 = 37
5 5 250 = 2 * 5^3
5 0 125 = 5^3
5 -2 117 = 3^2 * 13
5 -3 98 = 2 * 7^2
5 -4 61 = 61
6 -2 208 = 2^4 * 13
7 5 468 = 2^2 * 3^2 * 13
7 3 370 = 2 * 5 * 37
7 -5 218 = 2 * 109
8 8 1024 = 2^10
8 5 637 = 7^2 * 13
8 4 576 = 2^6 * 3^2
8 2 520 = 2^3 * 5 * 13
8 0 512 = 2^9
8 -3 485 = 5 * 97
8 -6 296 = 2^3 * 37
8 -7 169 = 13^2
9 9 1458 = 2 * 3^6
9 8 1241 = 17 * 73
9 4 793 = 13 * 61
9 1 730 = 2 * 5 * 73
9 0 729 = 3^6
9 -7 386 = 2 * 193
10 10 2000 = 2^4 * 5^3
10 5 1125 = 3^2 * 5^3
10 0 1000 = 2^3 * 5^3
10 -4 936 = 2^3 * 3^2 * 13
10 -6 784 = 2^4 * 7^2
10 -7 657 = 3^2 * 73
10 -8 488 = 2^3 * 61
11 1 1332 = 2^2 * 3^2 * 37
12 5 1853 = 17 * 109
12 -4 1664 = 2^7 * 13
12 -7 1385 = 5 * 277
12 -11 397 = 397
13 13 4394 = 2 * 13^3
13 12 3925 = 5^2 * 157
13 11 3528 = 2^3 * 3^2 * 7^2
13 2 2205 = 3^2 * 5 * 7^2
13 0 2197 = 13^3
13 -1 2196 = 2^2 * 3^2 * 61
13 -8 1685 = 5 * 337
13 -11 866 = 2 * 433
14 13 4941 = 3^4 * 61
14 10 3744 = 2^5 * 3^2 * 13
14 6 2960 = 2^4 * 5 * 37
14 1 2745 = 3^2 * 5 * 61
14 -7 2401 = 7^4
14 -10 1744 = 2^4 * 109
14 -11 1413 = 3^2 * 157
15 11 4706 = 2 * 13 * 181
15 -5 3250 = 2 * 5^3 * 13
16 16 8192 = 2^13
16 10 5096 = 2^3 * 7^2 * 13
16 9 4825 = 5^2 * 193
16 8 4608 = 2^9 * 3^2
16 4 4160 = 2^6 * 5 * 13
16 1 4097 = 17 * 241
16 0 4096 = 2^12
16 -3 4069 = 13 * 313
16 -6 3880 = 2^3 * 5 * 97
16 -12 2368 = 2^6 * 37
16 -14 1352 = 2^3 * 13^2
17 17 9826 = 2 * 17^3
17 12 6641 = 29 * 229
17 10 5913 = 3^4 * 73
17 7 5256 = 2^3 * 3^2 * 73
17 0 4913 = 17^3
17 -2 4905 = 3^2 * 5 * 109
17 -4 4849 = 13 * 373
17 -7 4570 = 2 * 5 * 457
17 -12 3185 = 5 * 7^2 * 13
17 -14 2169 = 3^2 * 241
17 -15 1538 = 2 * 769
18 18 11664 = 2^4 * 3^6
18 16 9928 = 2^3 * 17 * 73
18 9 6561 = 3^8
18 8 6344 = 2^3 * 13 * 61
18 2 5840 = 2^4 * 5 * 73
18 0 5832 = 2^3 * 3^6
18 -14 3088 = 2^4 * 193
19 7 7202 = 2 * 13 * 277
19 5 6984 = 2^3 * 3^2 * 97
19 -7 6516 = 2^2 * 3^2 * 181
19 -9 6130 = 2 * 5 * 613
20 20 16000 = 2^7 * 5^3
20 17 12913 = 37 * 349
20 10 9000 = 2^3 * 3^2 * 5^3
20 5 8125 = 5^4 * 13
20 0 8000 = 2^6 * 5^3
20 -8 7488 = 2^6 * 3^2 * 13
20 -12 6272 = 2^7 * 7^2
20 -14 5256 = 2^3 * 3^2 * 73
20 -15 4625 = 5^3 * 37
20 -16 3904 = 2^6 * 61
jagy@phobeusjunior:~$
$a=x^3-3x^2y-3xy^2+y^3,b=x^3+3x^2y-3xy^2-y^3,c=d=x^2+y^2.$
$a=3(x^3-3xy^2),b=3(3x^2y-y^3),c=x^2+y^2,d=2(x^2+y^2)$
$\cdots$
See this post.
On
Let $c=x^2,d=y^2,i=\sqrt{-1}$, then $$c^3+d^3=x^6+y^6=(x^3-(yi)^3)(x^3+(yi)^3)\\ =(x-yi)(x^2+xyi-y^2)(x+yi)(x^2-xyi-y^2)$$
Let $a+bi=(x-yi)(x^2-xyi-y^2),a-bi=(x+yi)(x^2+xyi-y^2)$, then $a^2+b^2=c^3+d^3$.
We get $a=x^3-2xy^2,b=y^3-2x^2y,c=x^2,d=y^2.$
If $a<0$ or $b<0$, we can take the absolute value.
For example, let $x=1,y=2$ we get $a=-7,b=4,c=1,d=4$ hence $7^2+4^2=1^3+4^3,$
let $x=3,y=2,$ we get $a=3,b=-28,c=9,d=4$ hence $3^2+28^2=9^3+4^3.$
Equation:
$$x^2+y^2=z^3+u^3$$
Formula of the solution, you can write:
$$x=q^6-2(a+s+t)q^5+(t^2-2(4a+3s)t-3a^2-10as-s^2)q^4-$$ $$-4(3t^3+(5a+4s)t^2+(3a^2+2as+s^2)t+a(a^2-s^2))q^3+$$ $$+(7t^4+4(a+s)t^3+6(3a^2+2as+s^2)t^2+4(3a^3+9sa^2+3as^2-s^3)t+3a^4+12sa^3+$$ $$+18s^2a^2-4as^3-s^4)q^2-(t^2-2ts+a^2-2as-s^2)(10t^3+18(a+s)t^2+$$ $$+2(5a^2+8as+5s^2)t+2(a^3+sa^2+as^2+s^3))q+(t^2-2at-a^2-2as+s^2)(7t^4+$$ $$+10(a+s)t^3+8(a^2+as+s^2)t^2+2(a^3+sa^2+as^2+s^3)t+a^4+2a^2s^2+s^4)$$
$$..............................................................$$
$$y=q^6+2(a+s+t)q^5+(t^2-2(3a+4s)t-a^2-10as-3s^2)q^4+$$ $$+4(3t^3+(4a+5s)t^2+(a^2+2as+3s^2)t+s(s^2-a^2))q^3+$$ $$+(7t^4+4(a+s)t^3+6(a^2+2as+3s^2)t^2+4(-a^3+3sa^2+9as^2+3s^3)t-a^4-4sa^3+$$ $$+18a^2s^2+12as^3+3s^4)q^2+(t^2-2at-a^2-2as+s^2)(10t^3+18(a+s)t^2+$$ $$+2(5a^2+8as+5s^2)t+2(a^3+sa^2+as^2+s^3))q+(t^2-2ts+a^2-2as-s^2)(7t^4+$$ $$+10(a+s)t^3+8(a^2+as+s^2)t^2+2(a^3+sa^2+as^2+s^3)t+a^4+2a^2s^2+s^4)$$
$$.............................................................$$
$$z=q^4-2(t^2+a^2+s^2+4at+4as+4st)q^2-3t^4-8(a+s)t^3-$$ $$-2(a^2+4as+s^2)t^2+a^4+2a^2s^2+s^4$$
$$..............................................................$$
$$u=(q^2+t^2+a^2+s^2)(q^2+5t^2+4(a+s)t+a^2+s^2)$$
$q,a,s,t$ - integers of any sign.
After substitution and obtain numerical results. It should be divided into common divisor. To get a primitive solution.