Mixing
An analogue to Euler's sum of powers conjecture
$begingroup$
What if instead of considering $$a_1^3+a_2^3+a_3^3=b_1^3 \
a_1^4+a_2^4+a_3^4+a_4^4=b_1^4 \
vdots $$
we instead considered
$$a_1^3+a_2^3+a_3^3=b_1^3+b_2^3 \
a_1^4+a_2^4+a_3^4+a_4^4=b_1^4+b_2^4+b_3^4 \
a_1^5+a_2^5+a_3^5+a_4^5+a_5^5=b_1^5+b_2^5+b_3^5+b_4^5 \
vdots $$ My gut feeling is that this will have infinitely many solutions.
Note: $a_i,b_i$ are all positive integers.
elementary-number-theory
asked Jan 10 at 11:24
Stupid Questions IncStupid Questions Inc
7010
$endgroup$
add a comment |
$begingroup$
What if instead of considering $$a_1^3+a_2^3+a_3^3=b_1^3 \
a_1^4+a_2^4+a_3^4+a_4^4=b_1^4 \
vdots $$
we instead considered
$$a_1^3+a_2^3+a_3^3=b_1^3+b_2^3 \
a_1^4+a_2^4+a_3^4+a_4^4=b_1^4+b_2^4+b_3^4 \
a_1^5+a_2^5+a_3^5+a_4^5+a_5^5=b_1^5+b_2^5+b_3^5+b_4^5 \
vdots $$ My gut feeling is that this will have infinitely many solutions.
Note: $a_i,b_i$ are all positive integers.
elementary-number-theory
asked Jan 10 at 11:24
Stupid Questions IncStupid Questions Inc
7010
$endgroup$
add a comment |
$begingroup$
What if instead of considering $$a_1^3+a_2^3+a_3^3=b_1^3 \
a_1^4+a_2^4+a_3^4+a_4^4=b_1^4 \
vdots $$
we instead considered
$$a_1^3+a_2^3+a_3^3=b_1^3+b_2^3 \
a_1^4+a_2^4+a_3^4+a_4^4=b_1^4+b_2^4+b_3^4 \
a_1^5+a_2^5+a_3^5+a_4^5+a_5^5=b_1^5+b_2^5+b_3^5+b_4^5 \
vdots $$ My gut feeling is that this will have infinitely many solutions.
Note: $a_i,b_i$ are all positive integers.
elementary-number-theory
asked Jan 10 at 11:24
Stupid Questions IncStupid Questions Inc
7010
$endgroup$
What if instead of considering $$a_1^3+a_2^3+a_3^3=b_1^3 \
a_1^4+a_2^4+a_3^4+a_4^4=b_1^4 \
vdots $$
we instead considered
$$a_1^3+a_2^3+a_3^3=b_1^3+b_2^3 \
a_1^4+a_2^4+a_3^4+a_4^4=b_1^4+b_2^4+b_3^4 \
a_1^5+a_2^5+a_3^5+a_4^5+a_5^5=b_1^5+b_2^5+b_3^5+b_4^5 \
vdots $$ My gut feeling is that this will have infinitely many solutions.
Note: $a_i,b_i$ are all positive integers.
elementary-number-theory
elementary-number-theory
asked Jan 10 at 11:24
Stupid Questions IncStupid Questions Inc
7010
asked Jan 10 at 11:24
Stupid Questions IncStupid Questions Inc
7010
edited Jan 11 at 7:23
Stupid Questions Inc
asked Jan 10 at 11:24
Stupid Questions IncStupid Questions Inc
7010
asked Jan 10 at 11:24
Stupid Questions IncStupid Questions Inc
7010
asked Jan 10 at 11:24
Stupid Questions IncStupid Questions Inc
7010
7010
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Here's a possible strategy to show that there are infinitely many solutions for the equation. For example, let's consider the first one. Suppose that there exists $x, y, zin mathbb{Z}$ with $zneq pm 1$ and satisfies $x^{3} + y^{3} = z^{3} + 2$. Then this gives one parametrization of the solution as
$$
(z^{m})^{3} + (z^{m})^{3} + (z^{m+1})^{3} = z^{3m}(z^{3} + 2) = z^{3m}(x^{3} + y^{3}) = (xz^{m})^{3} + (yz^{m})^{3}
$$
for $mgeq 0$. Also, we have a solution for the previous equation as $7^{3} + (-5)^{3} = 6^{3} + 2$, so we just proved that the equation $a_{1}^{3} + a_{2}^{3} + a_{3}^{3} = b_{1}^{3} + b_{2}^{3}$ has infinitely many integer solutions.
If you only want positive solutions, it is also possible because we have
$$
1214928^{3} + 3480205^{3} = 3528875^{3} + 2.
$$
Similarly, it is enough to show that there exists an integer $1leq t leq k-1$ such that the equation
$$
x_{1}^{k} + x_{2}^{k} + cdots + x_{k-1}^{k} = ty^{k} + (k-t) = t(y^{k} - 1) + k
$$
has an integer (or positive integer) solution with $yneq pm 1$, which will give a parametrization
$$
(y^{m})^{k} + cdots + (y^{m})^{k} + (y^{m+1})^{k} + cdots + (y^{m+1})^{k} = (x_{1}y^{m})^{k} + cdots + (x_{k-1}y^{m})^{k}
$$
where there are $(k-t)$ many $y^{m}$'s and $t$ many $y^{m+1}$'s on LHS.
answered Jan 10 at 13:11
Seewoo LeeSeewoo Lee
6,616926
$endgroup$
$begingroup$
+1 for ingenuity, although it looks challenging, for higher powers, to find equations with suitable $t$.
$endgroup$
– Adam Bailey
Jan 10 at 22:43
$begingroup$
Does your strategy work if we restrict everything to the naturals?
$endgroup$
– Stupid Questions Inc
Jan 13 at 19:50
$begingroup$
@StupidQuestionsInc Sure, if you can find the solution of the last equation with all $x_{i}>0$ and $y>1$.
$endgroup$
– Seewoo Lee
Jan 13 at 22:06
add a comment |
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1 Answer
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active
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1 Answer
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active
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active
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$begingroup$
Here's a possible strategy to show that there are infinitely many solutions for the equation. For example, let's consider the first one. Suppose that there exists $x, y, zin mathbb{Z}$ with $zneq pm 1$ and satisfies $x^{3} + y^{3} = z^{3} + 2$. Then this gives one parametrization of the solution as
$$
(z^{m})^{3} + (z^{m})^{3} + (z^{m+1})^{3} = z^{3m}(z^{3} + 2) = z^{3m}(x^{3} + y^{3}) = (xz^{m})^{3} + (yz^{m})^{3}
$$
for $mgeq 0$. Also, we have a solution for the previous equation as $7^{3} + (-5)^{3} = 6^{3} + 2$, so we just proved that the equation $a_{1}^{3} + a_{2}^{3} + a_{3}^{3} = b_{1}^{3} + b_{2}^{3}$ has infinitely many integer solutions.
If you only want positive solutions, it is also possible because we have
$$
1214928^{3} + 3480205^{3} = 3528875^{3} + 2.
$$
Similarly, it is enough to show that there exists an integer $1leq t leq k-1$ such that the equation
$$
x_{1}^{k} + x_{2}^{k} + cdots + x_{k-1}^{k} = ty^{k} + (k-t) = t(y^{k} - 1) + k
$$
has an integer (or positive integer) solution with $yneq pm 1$, which will give a parametrization
$$
(y^{m})^{k} + cdots + (y^{m})^{k} + (y^{m+1})^{k} + cdots + (y^{m+1})^{k} = (x_{1}y^{m})^{k} + cdots + (x_{k-1}y^{m})^{k}
$$
where there are $(k-t)$ many $y^{m}$'s and $t$ many $y^{m+1}$'s on LHS.
answered Jan 10 at 13:11
Seewoo LeeSeewoo Lee
6,616926
$endgroup$
$begingroup$
+1 for ingenuity, although it looks challenging, for higher powers, to find equations with suitable $t$.
$endgroup$
– Adam Bailey
Jan 10 at 22:43
$begingroup$
Does your strategy work if we restrict everything to the naturals?
$endgroup$
– Stupid Questions Inc
Jan 13 at 19:50
$begingroup$
@StupidQuestionsInc Sure, if you can find the solution of the last equation with all $x_{i}>0$ and $y>1$.
$endgroup$
– Seewoo Lee
Jan 13 at 22:06
add a comment |
$begingroup$
Here's a possible strategy to show that there are infinitely many solutions for the equation. For example, let's consider the first one. Suppose that there exists $x, y, zin mathbb{Z}$ with $zneq pm 1$ and satisfies $x^{3} + y^{3} = z^{3} + 2$. Then this gives one parametrization of the solution as
$$
(z^{m})^{3} + (z^{m})^{3} + (z^{m+1})^{3} = z^{3m}(z^{3} + 2) = z^{3m}(x^{3} + y^{3}) = (xz^{m})^{3} + (yz^{m})^{3}
$$
for $mgeq 0$. Also, we have a solution for the previous equation as $7^{3} + (-5)^{3} = 6^{3} + 2$, so we just proved that the equation $a_{1}^{3} + a_{2}^{3} + a_{3}^{3} = b_{1}^{3} + b_{2}^{3}$ has infinitely many integer solutions.
If you only want positive solutions, it is also possible because we have
$$
1214928^{3} + 3480205^{3} = 3528875^{3} + 2.
$$
Similarly, it is enough to show that there exists an integer $1leq t leq k-1$ such that the equation
$$
x_{1}^{k} + x_{2}^{k} + cdots + x_{k-1}^{k} = ty^{k} + (k-t) = t(y^{k} - 1) + k
$$
has an integer (or positive integer) solution with $yneq pm 1$, which will give a parametrization
$$
(y^{m})^{k} + cdots + (y^{m})^{k} + (y^{m+1})^{k} + cdots + (y^{m+1})^{k} = (x_{1}y^{m})^{k} + cdots + (x_{k-1}y^{m})^{k}
$$
where there are $(k-t)$ many $y^{m}$'s and $t$ many $y^{m+1}$'s on LHS.
answered Jan 10 at 13:11
Seewoo LeeSeewoo Lee
6,616926
$endgroup$
$begingroup$
+1 for ingenuity, although it looks challenging, for higher powers, to find equations with suitable $t$.
$endgroup$
– Adam Bailey
Jan 10 at 22:43
$begingroup$
Does your strategy work if we restrict everything to the naturals?
$endgroup$
– Stupid Questions Inc
Jan 13 at 19:50
$begingroup$
@StupidQuestionsInc Sure, if you can find the solution of the last equation with all $x_{i}>0$ and $y>1$.
$endgroup$
– Seewoo Lee
Jan 13 at 22:06
add a comment |
$begingroup$
Here's a possible strategy to show that there are infinitely many solutions for the equation. For example, let's consider the first one. Suppose that there exists $x, y, zin mathbb{Z}$ with $zneq pm 1$ and satisfies $x^{3} + y^{3} = z^{3} + 2$. Then this gives one parametrization of the solution as
$$
(z^{m})^{3} + (z^{m})^{3} + (z^{m+1})^{3} = z^{3m}(z^{3} + 2) = z^{3m}(x^{3} + y^{3}) = (xz^{m})^{3} + (yz^{m})^{3}
$$
for $mgeq 0$. Also, we have a solution for the previous equation as $7^{3} + (-5)^{3} = 6^{3} + 2$, so we just proved that the equation $a_{1}^{3} + a_{2}^{3} + a_{3}^{3} = b_{1}^{3} + b_{2}^{3}$ has infinitely many integer solutions.
If you only want positive solutions, it is also possible because we have
$$
1214928^{3} + 3480205^{3} = 3528875^{3} + 2.
$$
Similarly, it is enough to show that there exists an integer $1leq t leq k-1$ such that the equation
$$
x_{1}^{k} + x_{2}^{k} + cdots + x_{k-1}^{k} = ty^{k} + (k-t) = t(y^{k} - 1) + k
$$
has an integer (or positive integer) solution with $yneq pm 1$, which will give a parametrization
$$
(y^{m})^{k} + cdots + (y^{m})^{k} + (y^{m+1})^{k} + cdots + (y^{m+1})^{k} = (x_{1}y^{m})^{k} + cdots + (x_{k-1}y^{m})^{k}
$$
where there are $(k-t)$ many $y^{m}$'s and $t$ many $y^{m+1}$'s on LHS.
answered Jan 10 at 13:11
Seewoo LeeSeewoo Lee
6,616926
$endgroup$
Here's a possible strategy to show that there are infinitely many solutions for the equation. For example, let's consider the first one. Suppose that there exists $x, y, zin mathbb{Z}$ with $zneq pm 1$ and satisfies $x^{3} + y^{3} = z^{3} + 2$. Then this gives one parametrization of the solution as
$$
(z^{m})^{3} + (z^{m})^{3} + (z^{m+1})^{3} = z^{3m}(z^{3} + 2) = z^{3m}(x^{3} + y^{3}) = (xz^{m})^{3} + (yz^{m})^{3}
$$
for $mgeq 0$. Also, we have a solution for the previous equation as $7^{3} + (-5)^{3} = 6^{3} + 2$, so we just proved that the equation $a_{1}^{3} + a_{2}^{3} + a_{3}^{3} = b_{1}^{3} + b_{2}^{3}$ has infinitely many integer solutions.
If you only want positive solutions, it is also possible because we have
$$
1214928^{3} + 3480205^{3} = 3528875^{3} + 2.
$$
Similarly, it is enough to show that there exists an integer $1leq t leq k-1$ such that the equation
$$
x_{1}^{k} + x_{2}^{k} + cdots + x_{k-1}^{k} = ty^{k} + (k-t) = t(y^{k} - 1) + k
$$
has an integer (or positive integer) solution with $yneq pm 1$, which will give a parametrization
$$
(y^{m})^{k} + cdots + (y^{m})^{k} + (y^{m+1})^{k} + cdots + (y^{m+1})^{k} = (x_{1}y^{m})^{k} + cdots + (x_{k-1}y^{m})^{k}
$$
where there are $(k-t)$ many $y^{m}$'s and $t$ many $y^{m+1}$'s on LHS.
answered Jan 10 at 13:11
Seewoo LeeSeewoo Lee
6,616926
answered Jan 10 at 13:11
Seewoo LeeSeewoo Lee
6,616926
answered Jan 10 at 13:11
Seewoo LeeSeewoo Lee
6,616926
answered Jan 10 at 13:11
Seewoo LeeSeewoo Lee
6,616926
6,616926
$begingroup$
+1 for ingenuity, although it looks challenging, for higher powers, to find equations with suitable $t$.
$endgroup$
– Adam Bailey
Jan 10 at 22:43
$begingroup$
Does your strategy work if we restrict everything to the naturals?
$endgroup$
– Stupid Questions Inc
Jan 13 at 19:50
$begingroup$
@StupidQuestionsInc Sure, if you can find the solution of the last equation with all $x_{i}>0$ and $y>1$.
$endgroup$
– Seewoo Lee
Jan 13 at 22:06
add a comment |
$begingroup$
+1 for ingenuity, although it looks challenging, for higher powers, to find equations with suitable $t$.
$endgroup$
– Adam Bailey
Jan 10 at 22:43
$begingroup$
Does your strategy work if we restrict everything to the naturals?
$endgroup$
– Stupid Questions Inc
Jan 13 at 19:50
$begingroup$
@StupidQuestionsInc Sure, if you can find the solution of the last equation with all $x_{i}>0$ and $y>1$.
$endgroup$
– Seewoo Lee
Jan 13 at 22:06
$begingroup$
+1 for ingenuity, although it looks challenging, for higher powers, to find equations with suitable $t$.
$endgroup$
– Adam Bailey
Jan 10 at 22:43
$begingroup$
+1 for ingenuity, although it looks challenging, for higher powers, to find equations with suitable $t$.
$endgroup$
– Adam Bailey
Jan 10 at 22:43
$begingroup$
Does your strategy work if we restrict everything to the naturals?
$endgroup$
– Stupid Questions Inc
Jan 13 at 19:50
$begingroup$
Does your strategy work if we restrict everything to the naturals?
$endgroup$
– Stupid Questions Inc
Jan 13 at 19:50
$begingroup$
@StupidQuestionsInc Sure, if you can find the solution of the last equation with all $x_{i}>0$ and $y>1$.
$endgroup$
– Seewoo Lee
Jan 13 at 22:06
$begingroup$
@StupidQuestionsInc Sure, if you can find the solution of the last equation with all $x_{i}>0$ and $y>1$.
$endgroup$
– Seewoo Lee
Jan 13 at 22:06
add a comment |
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