Find all positive integers a,b,c,d such that $2019^a = b^3+c^3+d^3-5$
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Working on a problem...
Find all positive integers a,b,c,d such that $2019^a = b^3+c^3+d^3-5$
Using $(a+b+c)^3$ yields far too many terms, and I cannot come up with any solutions other than guessing and checking.
contest-math
asked Jan 10 at 4:56
Hubert LioniHubert Lioni
233
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add a comment |
$begingroup$
Working on a problem...
Find all positive integers a,b,c,d such that $2019^a = b^3+c^3+d^3-5$
Using $(a+b+c)^3$ yields far too many terms, and I cannot come up with any solutions other than guessing and checking.
contest-math
asked Jan 10 at 4:56
Hubert LioniHubert Lioni
233
$endgroup$
add a comment |
$begingroup$
Working on a problem...
Find all positive integers a,b,c,d such that $2019^a = b^3+c^3+d^3-5$
Using $(a+b+c)^3$ yields far too many terms, and I cannot come up with any solutions other than guessing and checking.
contest-math
asked Jan 10 at 4:56
Hubert LioniHubert Lioni
233
$endgroup$
Working on a problem...
Find all positive integers a,b,c,d such that $2019^a = b^3+c^3+d^3-5$
Using $(a+b+c)^3$ yields far too many terms, and I cannot come up with any solutions other than guessing and checking.
contest-math
contest-math
asked Jan 10 at 4:56
Hubert LioniHubert Lioni
233
asked Jan 10 at 4:56
Hubert LioniHubert Lioni
233
asked Jan 10 at 4:56
Hubert LioniHubert Lioni
233
asked Jan 10 at 4:56
Hubert LioniHubert Lioni
233
asked Jan 10 at 4:56
Hubert LioniHubert Lioni
233
233
add a comment |
add a comment |
1 Answer
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HINT: If $a ge 2$, then $2019^a equiv 0 pmod{9}$, while $b^3 equiv -1, 0, text{or} 1 pmod{9}$ and similarly for $c^3$ and $d^3$.
Do you see a contradiction? After ruling out $a ge 2$, you only need to solve the $a = 1$ case, i.e. $b^3+c^3+d^3 = 2024$. Since $b,c,d$ are positive integers, none of them can be larger than $sqrt[3]{2024}$, so there aren't too many possibilities.
answered Jan 10 at 5:18
JimmyK4542JimmyK4542
41.2k245107
$endgroup$
1
$begingroup$
Interesting, I solved a combination (10,8,8) earlier and wondered if this triple (and its permutations) are the only solutions. By your logic, they should be.
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– Hubert Lioni
Jan 10 at 6:05
add a comment |
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1 Answer
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$begingroup$
HINT: If $a ge 2$, then $2019^a equiv 0 pmod{9}$, while $b^3 equiv -1, 0, text{or} 1 pmod{9}$ and similarly for $c^3$ and $d^3$.
Do you see a contradiction? After ruling out $a ge 2$, you only need to solve the $a = 1$ case, i.e. $b^3+c^3+d^3 = 2024$. Since $b,c,d$ are positive integers, none of them can be larger than $sqrt[3]{2024}$, so there aren't too many possibilities.
answered Jan 10 at 5:18
JimmyK4542JimmyK4542
41.2k245107
$endgroup$
1
$begingroup$
Interesting, I solved a combination (10,8,8) earlier and wondered if this triple (and its permutations) are the only solutions. By your logic, they should be.
$endgroup$
– Hubert Lioni
Jan 10 at 6:05
add a comment |
$begingroup$
HINT: If $a ge 2$, then $2019^a equiv 0 pmod{9}$, while $b^3 equiv -1, 0, text{or} 1 pmod{9}$ and similarly for $c^3$ and $d^3$.
Do you see a contradiction? After ruling out $a ge 2$, you only need to solve the $a = 1$ case, i.e. $b^3+c^3+d^3 = 2024$. Since $b,c,d$ are positive integers, none of them can be larger than $sqrt[3]{2024}$, so there aren't too many possibilities.
answered Jan 10 at 5:18
JimmyK4542JimmyK4542
41.2k245107
$endgroup$
1
$begingroup$
Interesting, I solved a combination (10,8,8) earlier and wondered if this triple (and its permutations) are the only solutions. By your logic, they should be.
$endgroup$
– Hubert Lioni
Jan 10 at 6:05
add a comment |
$begingroup$
HINT: If $a ge 2$, then $2019^a equiv 0 pmod{9}$, while $b^3 equiv -1, 0, text{or} 1 pmod{9}$ and similarly for $c^3$ and $d^3$.
Do you see a contradiction? After ruling out $a ge 2$, you only need to solve the $a = 1$ case, i.e. $b^3+c^3+d^3 = 2024$. Since $b,c,d$ are positive integers, none of them can be larger than $sqrt[3]{2024}$, so there aren't too many possibilities.
answered Jan 10 at 5:18
JimmyK4542JimmyK4542
41.2k245107
$endgroup$
HINT: If $a ge 2$, then $2019^a equiv 0 pmod{9}$, while $b^3 equiv -1, 0, text{or} 1 pmod{9}$ and similarly for $c^3$ and $d^3$.
Do you see a contradiction? After ruling out $a ge 2$, you only need to solve the $a = 1$ case, i.e. $b^3+c^3+d^3 = 2024$. Since $b,c,d$ are positive integers, none of them can be larger than $sqrt[3]{2024}$, so there aren't too many possibilities.
answered Jan 10 at 5:18
JimmyK4542JimmyK4542
41.2k245107
answered Jan 10 at 5:18
JimmyK4542JimmyK4542
41.2k245107
answered Jan 10 at 5:18
JimmyK4542JimmyK4542
41.2k245107
answered Jan 10 at 5:18
JimmyK4542JimmyK4542
41.2k245107
41.2k245107
1
$begingroup$
Interesting, I solved a combination (10,8,8) earlier and wondered if this triple (and its permutations) are the only solutions. By your logic, they should be.
$endgroup$
– Hubert Lioni
Jan 10 at 6:05
add a comment |
1
$begingroup$
Interesting, I solved a combination (10,8,8) earlier and wondered if this triple (and its permutations) are the only solutions. By your logic, they should be.
$endgroup$
– Hubert Lioni
Jan 10 at 6:05
1
1
$begingroup$
Interesting, I solved a combination (10,8,8) earlier and wondered if this triple (and its permutations) are the only solutions. By your logic, they should be.
$endgroup$
– Hubert Lioni
Jan 10 at 6:05
$begingroup$
Interesting, I solved a combination (10,8,8) earlier and wondered if this triple (and its permutations) are the only solutions. By your logic, they should be.
$endgroup$
– Hubert Lioni
Jan 10 at 6:05
add a comment |
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