-4
$\begingroup$

Find all the integer solutions $a, b, c, d, e, f$ satisfying the equation $a^2b^3 + c^2d^3 = e^2f^3$.

Note that if we prove that there are no such solutions with the condition $\text{gcd}(ab, cd, ef) = 1$ and $|bdf|>1$ then we get simpler proof of Fermat's last theorem, because each positive integer $n>1$ has a representation $n=2k+3l$.

Improve this question
$\endgroup$
5
  • $\begingroup$ As you say, this is harder than proving the modularity of elliptic curves. $\endgroup$
    – ThiKu
    Commented Mar 26, 2019 at 8:15
  • 1
    $\begingroup$ $a=2,b=1,c=11,d=1,e=1,f=5$ $\endgroup$
    – Wojowu
    Commented Mar 26, 2019 at 8:21
  • 1
    $\begingroup$ $100+243=343$ ($a,b,c,d,e,f=10,1,3,3,1,7$). $1+675=676$. $\endgroup$
    – Gerry Myerson
    Commented Mar 26, 2019 at 9:03
  • 1
    $\begingroup$ $3125+4563=7688$ ($a=5,b=5,c=13,d=3,e=31,f=2$). $\endgroup$
    – Robert Israel
    Commented Mar 26, 2019 at 14:31
  • $\begingroup$ @ThiKu I think you do not really need full modularity of elliptic curves over Q for Fermat's theorem; I think it was proved after Fermat's theorem has been proved. Or do you mean something else? $\endgroup$
    – user74900
    Commented Mar 26, 2019 at 14:46

1 Answer 1

Reset to default
1
$\begingroup$

Some solutions with $a,b,c,d,e,f>1$: $$ \eqalign{3125 + 4563 &= 5^3 \cdot 5^2 + 3^3\cdot 13^2 = 7688 = 2^3\cdot 31^2\cr 4563 + 66125 &= 3^3 \cdot 13^2 + 5^3\cdot 23^2 = 70688 = 2^3\cdot 94^2\cr 45125 + 49923 &= 5^3 \cdot 19^2 + 3^3\cdot 43^2 = 95048 = 2^3\cdot 109^2\cr 19773 + 100352 &= 13^3 \cdot 3^2 + 8^3\cdot 14^2 = 120125 = 5^3\cdot 31^2\cr 25947 + 105125 &= 3^3 \cdot 31^2 + 5^3\cdot 29^2 = 131072 = 32^3\cdot 2^2\cr 41503 + 144500 &= 7^3 \cdot 11^2 + 5^3\cdot 34^2 = 186003 = 3^3\cdot 83^2\cr 15125 + 204363 &= 5^3 \cdot 11^2 + 3^3\cdot 87^2 = 219488 = 38^3\cdot 2^2\cr 78125 + 141376 &= 5^3 \cdot 25^2 + 4^3\cdot 47^2 = 219501 = 29^3\cdot 3^2\cr 66125 + 199712 &= 5^3 \cdot 23^2 + 2^3\cdot 158^2 = 265837 = 13^3\cdot 11^2\cr 61504 + 224939 &= 4^3 \cdot 31^2 + 11^3\cdot 13^2 = 286443 = 3^3\cdot 103^2\cr 3087 + 301088 &= 7^3 \cdot 3^2 + 2^3\cdot 194^2 = 304175 = 23^3\cdot 5^2\cr 15125 + 356168 &= 5^3 \cdot 11^2 + 2^3\cdot 211^2 = 371293 = 13^3\cdot 13^2\cr 136107 + 276125 &= 3^3 \cdot 71^2 + 5^3\cdot 47^2 = 412232 = 2^3\cdot 227^2\cr 231125 + 233523 &= 5^3 \cdot 43^2 + 3^3\cdot 93^2 = 464648 = 2^3\cdot 241^2\cr 9248 + 455877 &= 2^3 \cdot 34^2 + 37^3\cdot 3^2 = 465125 = 5^3\cdot 61^2\cr 675 + 595508 &= 3^3 \cdot 5^2 + 53^3\cdot 2^2 = 596183 = 23^3\cdot 7^2\cr 8575 + 620289 &= 7^3 \cdot 5^2 + 41^3\cdot 3^2 = 628864 = 34^3\cdot 4^2\cr 171475 + 528392 &= 19^3 \cdot 5^2 + 2^3\cdot 257^2 = 699867 = 3^3\cdot 161^2\cr 168507 + 561125 &= 3^3 \cdot 79^2 + 5^3\cdot 67^2 = 729632 = 2^3\cdot 302^2\cr 263169 + 469567 &= 9^3 \cdot 19^2 + 7^3\cdot 37^2 = 732736 = 4^3\cdot 107^2\cr 168507 + 666125 &= 3^3 \cdot 79^2 + 5^3\cdot 73^2 = 834632 = 2^3\cdot 323^2\cr 30375 + 830297 &= 15^3 \cdot 3^2 + 17^3\cdot 13^2 = 860672 = 8^3\cdot 41^2\cr }$$

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .