Let be $a, b, c, d$ positive integers. Consider the following equation:
$$a=\frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$$
$(a,b,c,d)=(1,1,1,1)$ is a solution $(a,b,c,d)=(5,4,6,1)$ is another solution.
Are there integer solutions with $d>1$?
2026-05-16 23:20:05.1778973605
On
On
Are there solutions to $a=\frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$ in positive integers $(a,b,c,d)$ with $d>1$?
207 Views Asked by user631773 https://math.techqa.club/user/user631773/detail At
4
There are 4 best solutions below
0
On
Let $b=c=d=t$. The expression simplifies to $a=t$. Hence, you can get many solutions where $d>1$.
0
On
I get it now. Begin with any triple such that $\gcd(b,c,d) = 1.$ The fraction is not necessarily an integer. However, there is a smallest multiplier, call it $k,$ such that using $(kb,kc,kd)$ does give integer $a,$ also $\gcd(a, kb,kc,kd) = 1.$ Also, the algorithm to find $k$ is easy. $$ k = \frac{2b^2+2c^2+d^2}{\gcd \left( 2b^2+2c^2+d^2 \;, \;2b^3+2c^3+d^3 \right)} $$
In particular, the mistaken triple (6,4,1) must be multiplied by 35 to get $a$ integral.
original 6 4 1 mult 35 gives 187 210 140 35
=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=
jagy@phobeusjunior:~$ ./mse
original 1 1 1 mult 1 gives 1 1 1 1
original 1 1 2 mult 2 gives 3 2 2 4
original 1 1 3 mult 13 gives 31 13 13 39
original 1 1 4 mult 5 gives 17 5 5 20
original 1 1 5 mult 29 gives 129 29 29 145
original 1 1 6 mult 2 gives 11 2 2 12
original 2 1 1 mult 11 gives 19 22 11 11
original 2 1 2 mult 7 gives 13 14 7 14
original 2 1 3 mult 19 gives 45 38 19 57
original 2 1 4 mult 13 gives 41 26 13 52
original 2 1 5 mult 35 gives 143 70 35 175
original 2 1 6 mult 23 gives 117 46 23 138
original 2 2 1 mult 17 gives 33 34 34 17
original 2 2 3 mult 25 gives 59 50 50 75
original 2 2 5 mult 41 gives 157 82 82 205
original 3 1 1 mult 7 gives 19 21 7 7
original 3 1 2 mult 3 gives 8 9 3 6
original 3 1 3 mult 29 gives 83 87 29 87
original 3 1 4 mult 3 gives 10 9 3 12
original 3 1 5 mult 45 gives 181 135 45 225
original 3 1 6 mult 7 gives 34 21 7 42
original 3 2 1 mult 27 gives 71 81 54 27
original 3 2 2 mult 5 gives 13 15 10 10
original 3 2 3 mult 35 gives 97 105 70 105
original 3 2 4 mult 21 gives 67 63 42 84
original 3 2 5 mult 17 gives 65 51 34 85
original 3 2 6 mult 31 gives 143 93 62 186
original 3 3 1 mult 37 gives 109 111 111 37
original 3 3 2 mult 10 gives 29 30 30 20
original 3 3 4 mult 13 gives 43 39 39 52
original 3 3 5 mult 61 gives 233 183 183 305
original 4 1 1 mult 35 gives 131 140 35 35
original 4 1 2 mult 19 gives 69 76 19 38
original 4 1 3 mult 43 gives 157 172 43 129
original 4 1 4 mult 25 gives 97 100 25 100
original 4 1 5 mult 59 gives 255 236 59 295
original 4 1 6 mult 35 gives 173 140 35 210
original 4 2 1 mult 41 gives 145 164 82 41
original 4 2 3 mult 49 gives 171 196 98 147
original 4 2 5 mult 65 gives 269 260 130 325
original 4 3 1 mult 17 gives 61 68 51 17
original 4 3 2 mult 27 gives 95 108 81 54
original 4 3 3 mult 59 gives 209 236 177 177
original 4 3 4 mult 11 gives 41 44 33 44
original 4 3 5 mult 75 gives 307 300 225 375
original 4 3 6 mult 43 gives 199 172 129 258
original 4 4 1 mult 65 gives 257 260 260 65
original 4 4 3 mult 73 gives 283 292 292 219
original 4 4 5 mult 89 gives 381 356 356 445
original 5 1 1 mult 53 gives 253 265 53 53
original 5 1 2 mult 14 gives 65 70 14 28
original 5 1 3 mult 61 gives 279 305 61 183
original 5 1 4 mult 17 gives 79 85 17 68
original 5 1 5 mult 77 gives 377 385 77 385
original 5 1 6 mult 22 gives 117 110 22 132
original 5 2 1 mult 59 gives 267 295 118 59
original 5 2 2 mult 31 gives 137 155 62 62
original 5 2 3 mult 67 gives 293 335 134 201
original 5 2 4 mult 37 gives 165 185 74 148
original 5 2 5 mult 83 gives 391 415 166 415
original 5 2 6 mult 47 gives 241 235 94 282
original 5 3 1 mult 69 gives 305 345 207 69
original 5 3 2 mult 3 gives 13 15 9 6
original 5 3 3 mult 77 gives 331 385 231 231
original 5 3 4 mult 21 gives 92 105 63 84
original 5 3 5 mult 31 gives 143 155 93 155
original 5 3 6 mult 1 gives 5 5 3 6
original 5 4 1 mult 83 gives 379 415 332 83
original 5 4 2 mult 43 gives 193 215 172 86
original 5 4 3 mult 91 gives 405 455 364 273
original 5 4 4 mult 49 gives 221 245 196 196
original 5 4 5 mult 107 gives 503 535 428 535
original 5 4 6 mult 59 gives 297 295 236 354
original 5 5 1 mult 101 gives 501 505 505 101
original 5 5 2 mult 26 gives 127 130 130 52
original 5 5 3 mult 109 gives 527 545 545 327
original 5 5 4 mult 29 gives 141 145 145 116
original 5 5 6 mult 34 gives 179 170 170 204
original 6 1 1 mult 5 gives 29 30 5 5
original 6 1 2 mult 3 gives 17 18 3 6
original 6 1 3 mult 83 gives 461 498 83 249
original 6 1 4 mult 15 gives 83 90 15 60
original 6 1 5 mult 99 gives 559 594 99 495
original 6 1 6 mult 11 gives 65 66 11 66
original 6 2 1 mult 81 gives 449 486 162 81
original 6 2 3 mult 89 gives 475 534 178 267
original 6 2 5 mult 35 gives 191 210 70 175
original 6 3 1 mult 91 gives 487 546 273 91
original 6 3 2 mult 47 gives 247 282 141 94
original 6 3 4 mult 53 gives 275 318 159 212
original 6 3 5 mult 115 gives 611 690 345 575
original 6 4 1 mult 35 gives 187 210 140 35
original 6 4 3 mult 113 gives 587 678 452 339
original 6 4 5 mult 129 gives 685 774 516 645
original 6 5 1 mult 123 gives 683 738 615 123
original 6 5 2 mult 21 gives 115 126 105 42
original 6 5 3 mult 131 gives 709 786 655 393
original 6 5 4 mult 69 gives 373 414 345 276
original 6 5 5 mult 49 gives 269 294 245 245
original 6 5 6 mult 79 gives 449 474 395 474
original 6 6 1 mult 29 gives 173 174 174 29
original 6 6 5 mult 169 gives 989 1014 1014 845
It seems that $\frac{2b^3+2c^3+d^3}{2b^2+2c^2+d^2}$ is not equal to 5 when $b=4$, $c=6$ and $d=1$, no ?
Brute-forcing can help you to get some solutions :