Mixing
Integer solutions to $t^p-(r^p+s^p)=pk_1$ and $t^p-left[(t-r)^p+(t-s)^pright]=p k_2$, where p is an odd prime
$begingroup$
This much re-edited question develops on from this question If $t^p-(r^p+s^p) equiv 0 (mod p)$ implies $t-(r+s)=pq$, do other solutions exist for relatively prime $r$, $s$ and $t$?
If the set of solutions to $t-(r+s) =pq$, where $q$ is a non-zero positive integer, contains the complete set of solutions in $(r,s,t,p)$ to $t^p-(r^p+s^p) equiv 0 ;( text{mod}; p)$
Then this implies that
$$t-(r+s) =pq tag{1}$$
and
$$t-((t-r)+(t-s))=(r+s)-t=pqtag{2}$$
, where $q$ is a positive integer, contain the complete set of solutions in $(r,s,t,p)$ to
$$t^p-(r^p+s^p)=p k_1tag{3}$$
and
$$t^p-left[(t-r)^p+(t-s)^pright]=p k_2tag{4}$$
respectively, where $k_1$ and $k_2$ are positive integers.
I think the sum of (3) and (4) gives
$$left(t^p-(r^p+s^p)right)+left(t^p-left[(t-r)^p+(t-s)^pright]right)=ptk=p t (frac{k_1+k_2}{t})$$
where $k_1+k_2$ is always divisible by t.
For the case $p=3$ with $t-(r+s)=3q$, we have
$$t^3-(r^3+s^3)=3left(3tq(r+s)+rst-3q(rs-3q^2)right)$$
and
$$t^3-left[(t-r)^3+(t-s)^3right]=3left(rst+3q(rs-3q^2)right)$$
with the sum of these being
$$left(t^3-(r^3+s^3)right)+left(t^3-left[(t-r)^3+(t-s)^3right]right)=3t(3q(r+s)+2rs)$$
If there exists a solution $t^3-left[(t-r)^3+(t-s)^3 right]=0$ then $3rst=-3^2q(rs-3q^2)$.
Now I have to work out if the algebra is tractable for any odd prime p.
elementary-number-theory
asked Jan 19 at 2:15
James ArathoonJames Arathoon
1,546423
$endgroup$
add a comment |
$begingroup$
This much re-edited question develops on from this question If $t^p-(r^p+s^p) equiv 0 (mod p)$ implies $t-(r+s)=pq$, do other solutions exist for relatively prime $r$, $s$ and $t$?
If the set of solutions to $t-(r+s) =pq$, where $q$ is a non-zero positive integer, contains the complete set of solutions in $(r,s,t,p)$ to $t^p-(r^p+s^p) equiv 0 ;( text{mod}; p)$
Then this implies that
$$t-(r+s) =pq tag{1}$$
and
$$t-((t-r)+(t-s))=(r+s)-t=pqtag{2}$$
, where $q$ is a positive integer, contain the complete set of solutions in $(r,s,t,p)$ to
$$t^p-(r^p+s^p)=p k_1tag{3}$$
and
$$t^p-left[(t-r)^p+(t-s)^pright]=p k_2tag{4}$$
respectively, where $k_1$ and $k_2$ are positive integers.
I think the sum of (3) and (4) gives
$$left(t^p-(r^p+s^p)right)+left(t^p-left[(t-r)^p+(t-s)^pright]right)=ptk=p t (frac{k_1+k_2}{t})$$
where $k_1+k_2$ is always divisible by t.
For the case $p=3$ with $t-(r+s)=3q$, we have
$$t^3-(r^3+s^3)=3left(3tq(r+s)+rst-3q(rs-3q^2)right)$$
and
$$t^3-left[(t-r)^3+(t-s)^3right]=3left(rst+3q(rs-3q^2)right)$$
with the sum of these being
$$left(t^3-(r^3+s^3)right)+left(t^3-left[(t-r)^3+(t-s)^3right]right)=3t(3q(r+s)+2rs)$$
If there exists a solution $t^3-left[(t-r)^3+(t-s)^3 right]=0$ then $3rst=-3^2q(rs-3q^2)$.
Now I have to work out if the algebra is tractable for any odd prime p.
elementary-number-theory
asked Jan 19 at 2:15
James ArathoonJames Arathoon
1,546423
$endgroup$
add a comment |
$begingroup$
This much re-edited question develops on from this question If $t^p-(r^p+s^p) equiv 0 (mod p)$ implies $t-(r+s)=pq$, do other solutions exist for relatively prime $r$, $s$ and $t$?
If the set of solutions to $t-(r+s) =pq$, where $q$ is a non-zero positive integer, contains the complete set of solutions in $(r,s,t,p)$ to $t^p-(r^p+s^p) equiv 0 ;( text{mod}; p)$
Then this implies that
$$t-(r+s) =pq tag{1}$$
and
$$t-((t-r)+(t-s))=(r+s)-t=pqtag{2}$$
, where $q$ is a positive integer, contain the complete set of solutions in $(r,s,t,p)$ to
$$t^p-(r^p+s^p)=p k_1tag{3}$$
and
$$t^p-left[(t-r)^p+(t-s)^pright]=p k_2tag{4}$$
respectively, where $k_1$ and $k_2$ are positive integers.
I think the sum of (3) and (4) gives
$$left(t^p-(r^p+s^p)right)+left(t^p-left[(t-r)^p+(t-s)^pright]right)=ptk=p t (frac{k_1+k_2}{t})$$
where $k_1+k_2$ is always divisible by t.
For the case $p=3$ with $t-(r+s)=3q$, we have
$$t^3-(r^3+s^3)=3left(3tq(r+s)+rst-3q(rs-3q^2)right)$$
and
$$t^3-left[(t-r)^3+(t-s)^3right]=3left(rst+3q(rs-3q^2)right)$$
with the sum of these being
$$left(t^3-(r^3+s^3)right)+left(t^3-left[(t-r)^3+(t-s)^3right]right)=3t(3q(r+s)+2rs)$$
If there exists a solution $t^3-left[(t-r)^3+(t-s)^3 right]=0$ then $3rst=-3^2q(rs-3q^2)$.
Now I have to work out if the algebra is tractable for any odd prime p.
elementary-number-theory
asked Jan 19 at 2:15
James ArathoonJames Arathoon
1,546423
$endgroup$
This much re-edited question develops on from this question If $t^p-(r^p+s^p) equiv 0 (mod p)$ implies $t-(r+s)=pq$, do other solutions exist for relatively prime $r$, $s$ and $t$?
If the set of solutions to $t-(r+s) =pq$, where $q$ is a non-zero positive integer, contains the complete set of solutions in $(r,s,t,p)$ to $t^p-(r^p+s^p) equiv 0 ;( text{mod}; p)$
Then this implies that
$$t-(r+s) =pq tag{1}$$
and
$$t-((t-r)+(t-s))=(r+s)-t=pqtag{2}$$
, where $q$ is a positive integer, contain the complete set of solutions in $(r,s,t,p)$ to
$$t^p-(r^p+s^p)=p k_1tag{3}$$
and
$$t^p-left[(t-r)^p+(t-s)^pright]=p k_2tag{4}$$
respectively, where $k_1$ and $k_2$ are positive integers.
I think the sum of (3) and (4) gives
$$left(t^p-(r^p+s^p)right)+left(t^p-left[(t-r)^p+(t-s)^pright]right)=ptk=p t (frac{k_1+k_2}{t})$$
where $k_1+k_2$ is always divisible by t.
For the case $p=3$ with $t-(r+s)=3q$, we have
$$t^3-(r^3+s^3)=3left(3tq(r+s)+rst-3q(rs-3q^2)right)$$
and
$$t^3-left[(t-r)^3+(t-s)^3right]=3left(rst+3q(rs-3q^2)right)$$
with the sum of these being
$$left(t^3-(r^3+s^3)right)+left(t^3-left[(t-r)^3+(t-s)^3right]right)=3t(3q(r+s)+2rs)$$
If there exists a solution $t^3-left[(t-r)^3+(t-s)^3 right]=0$ then $3rst=-3^2q(rs-3q^2)$.
Now I have to work out if the algebra is tractable for any odd prime p.
elementary-number-theory
elementary-number-theory
asked Jan 19 at 2:15
James ArathoonJames Arathoon
1,546423
asked Jan 19 at 2:15
James ArathoonJames Arathoon
1,546423
edited Jan 20 at 17:04
James Arathoon
asked Jan 19 at 2:15
James ArathoonJames Arathoon
1,546423
asked Jan 19 at 2:15
James ArathoonJames Arathoon
1,546423
asked Jan 19 at 2:15
James ArathoonJames Arathoon
1,546423
1,546423
add a comment |
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