Fifth root of $x$ raised to three… [closed]
$begingroup$
Find the values for $x$ and $y$ if the following equations are true.
begin{align}
sqrt[5]x^3 &= y^2 - 2\
x - y &= 7y
end{align}
The values of $x$ and $y$ are positive integers not more than $100$, also please note this isn't a homework.
systems-of-equations diophantine-equations radicals
edited Jan 23 at 12:49
Harry Peter
5,47911439
asked Jan 21 at 13:37
Tanmay GajapatiTanmay Gajapati
1
$endgroup$
closed as off-topic by Eevee Trainer, Did, Abcd, Lord_Farin, Paul Frost Jan 21 at 18:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Did, Abcd, Lord_Farin, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Find the values for $x$ and $y$ if the following equations are true.
begin{align}
sqrt[5]x^3 &= y^2 - 2\
x - y &= 7y
end{align}
The values of $x$ and $y$ are positive integers not more than $100$, also please note this isn't a homework.
systems-of-equations diophantine-equations radicals
edited Jan 23 at 12:49
Harry Peter
5,47911439
asked Jan 21 at 13:37
Tanmay GajapatiTanmay Gajapati
1
$endgroup$
closed as off-topic by Eevee Trainer, Did, Abcd, Lord_Farin, Paul Frost Jan 21 at 18:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Did, Abcd, Lord_Farin, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Well if I am not mistaken the second equation implies $y=frac x8$. Thus, you are left with an equation in $x$ alone, namely $$sqrt[5]{x}^3=frac{x^2}{64}-2$$
$endgroup$
– mrtaurho
Jan 21 at 13:42
add a comment |
$begingroup$
Find the values for $x$ and $y$ if the following equations are true.
begin{align}
sqrt[5]x^3 &= y^2 - 2\
x - y &= 7y
end{align}
The values of $x$ and $y$ are positive integers not more than $100$, also please note this isn't a homework.
systems-of-equations diophantine-equations radicals
edited Jan 23 at 12:49
Harry Peter
5,47911439
asked Jan 21 at 13:37
Tanmay GajapatiTanmay Gajapati
1
$endgroup$
Find the values for $x$ and $y$ if the following equations are true.
begin{align}
sqrt[5]x^3 &= y^2 - 2\
x - y &= 7y
end{align}
The values of $x$ and $y$ are positive integers not more than $100$, also please note this isn't a homework.
systems-of-equations diophantine-equations radicals
systems-of-equations diophantine-equations radicals
edited Jan 23 at 12:49
Harry Peter
5,47911439
asked Jan 21 at 13:37
Tanmay GajapatiTanmay Gajapati
1
edited Jan 23 at 12:49
Harry Peter
5,47911439
asked Jan 21 at 13:37
Tanmay GajapatiTanmay Gajapati
1
edited Jan 23 at 12:49
Harry Peter
5,47911439
edited Jan 23 at 12:49
Harry Peter
5,47911439
edited Jan 23 at 12:49
Harry Peter
5,47911439
5,47911439
asked Jan 21 at 13:37
Tanmay GajapatiTanmay Gajapati
1
asked Jan 21 at 13:37
Tanmay GajapatiTanmay Gajapati
1
asked Jan 21 at 13:37
Tanmay GajapatiTanmay Gajapati
1
1
closed as off-topic by Eevee Trainer, Did, Abcd, Lord_Farin, Paul Frost Jan 21 at 18:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Did, Abcd, Lord_Farin, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Eevee Trainer, Did, Abcd, Lord_Farin, Paul Frost Jan 21 at 18:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Did, Abcd, Lord_Farin, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Well if I am not mistaken the second equation implies $y=frac x8$. Thus, you are left with an equation in $x$ alone, namely $$sqrt[5]{x}^3=frac{x^2}{64}-2$$
$endgroup$
– mrtaurho
Jan 21 at 13:42
add a comment |
1
$begingroup$
Well if I am not mistaken the second equation implies $y=frac x8$. Thus, you are left with an equation in $x$ alone, namely $$sqrt[5]{x}^3=frac{x^2}{64}-2$$
$endgroup$
– mrtaurho
Jan 21 at 13:42
1
1
$begingroup$
Well if I am not mistaken the second equation implies $y=frac x8$. Thus, you are left with an equation in $x$ alone, namely $$sqrt[5]{x}^3=frac{x^2}{64}-2$$
$endgroup$
– mrtaurho
Jan 21 at 13:42
$begingroup$
Well if I am not mistaken the second equation implies $y=frac x8$. Thus, you are left with an equation in $x$ alone, namely $$sqrt[5]{x}^3=frac{x^2}{64}-2$$
$endgroup$
– mrtaurho
Jan 21 at 13:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
From the second equation $x=8y$. Putting this in first equation,
$$sqrt[5]{(8y)^3}=y^2-2$$
$$sqrt[5]{512y^3}=y^2-2 ldots(i)$$
Note that $y$ is an integer, so $y^2-2$ is also an integer. So $sqrt[5]{512y^3}$ is also an integer. This is only possible if in the prime factorization of $512y^3=2^9y^3$, the power of each factor is a multiple of $5$. By inspection, we can see that this is only possible if $y = 2^2k^5=4k^5$.
For $k=1$, $y=4$ and for $kge2$, $y>100$. So the only possible value of $y$ that fulfills the given conditions is $y=4$. However $y=4$ does not satisfy the equation $(i)$. So $y=4$ is also not admissible.
Hence there is no solution of the given equations such that $x, y in Z $ and $x,yle100$.
answered Jan 21 at 13:55
Faiq IrfanFaiq Irfan
814317
$endgroup$
$begingroup$
I tried to graph the two equations, I get two solutions (0,0) and (32,4) but I want an arithmetic explanation for it. Thank you.
$endgroup$
– Tanmay Gajapati
Jan 22 at 14:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From the second equation $x=8y$. Putting this in first equation,
$$sqrt[5]{(8y)^3}=y^2-2$$
$$sqrt[5]{512y^3}=y^2-2 ldots(i)$$
Note that $y$ is an integer, so $y^2-2$ is also an integer. So $sqrt[5]{512y^3}$ is also an integer. This is only possible if in the prime factorization of $512y^3=2^9y^3$, the power of each factor is a multiple of $5$. By inspection, we can see that this is only possible if $y = 2^2k^5=4k^5$.
For $k=1$, $y=4$ and for $kge2$, $y>100$. So the only possible value of $y$ that fulfills the given conditions is $y=4$. However $y=4$ does not satisfy the equation $(i)$. So $y=4$ is also not admissible.
Hence there is no solution of the given equations such that $x, y in Z $ and $x,yle100$.
answered Jan 21 at 13:55
Faiq IrfanFaiq Irfan
814317
$endgroup$
$begingroup$
I tried to graph the two equations, I get two solutions (0,0) and (32,4) but I want an arithmetic explanation for it. Thank you.
$endgroup$
– Tanmay Gajapati
Jan 22 at 14:48
add a comment |
$begingroup$
From the second equation $x=8y$. Putting this in first equation,
$$sqrt[5]{(8y)^3}=y^2-2$$
$$sqrt[5]{512y^3}=y^2-2 ldots(i)$$
Note that $y$ is an integer, so $y^2-2$ is also an integer. So $sqrt[5]{512y^3}$ is also an integer. This is only possible if in the prime factorization of $512y^3=2^9y^3$, the power of each factor is a multiple of $5$. By inspection, we can see that this is only possible if $y = 2^2k^5=4k^5$.
For $k=1$, $y=4$ and for $kge2$, $y>100$. So the only possible value of $y$ that fulfills the given conditions is $y=4$. However $y=4$ does not satisfy the equation $(i)$. So $y=4$ is also not admissible.
Hence there is no solution of the given equations such that $x, y in Z $ and $x,yle100$.
answered Jan 21 at 13:55
Faiq IrfanFaiq Irfan
814317
$endgroup$
$begingroup$
I tried to graph the two equations, I get two solutions (0,0) and (32,4) but I want an arithmetic explanation for it. Thank you.
$endgroup$
– Tanmay Gajapati
Jan 22 at 14:48
add a comment |
$begingroup$
From the second equation $x=8y$. Putting this in first equation,
$$sqrt[5]{(8y)^3}=y^2-2$$
$$sqrt[5]{512y^3}=y^2-2 ldots(i)$$
Note that $y$ is an integer, so $y^2-2$ is also an integer. So $sqrt[5]{512y^3}$ is also an integer. This is only possible if in the prime factorization of $512y^3=2^9y^3$, the power of each factor is a multiple of $5$. By inspection, we can see that this is only possible if $y = 2^2k^5=4k^5$.
For $k=1$, $y=4$ and for $kge2$, $y>100$. So the only possible value of $y$ that fulfills the given conditions is $y=4$. However $y=4$ does not satisfy the equation $(i)$. So $y=4$ is also not admissible.
Hence there is no solution of the given equations such that $x, y in Z $ and $x,yle100$.
answered Jan 21 at 13:55
Faiq IrfanFaiq Irfan
814317
$endgroup$
From the second equation $x=8y$. Putting this in first equation,
$$sqrt[5]{(8y)^3}=y^2-2$$
$$sqrt[5]{512y^3}=y^2-2 ldots(i)$$
Note that $y$ is an integer, so $y^2-2$ is also an integer. So $sqrt[5]{512y^3}$ is also an integer. This is only possible if in the prime factorization of $512y^3=2^9y^3$, the power of each factor is a multiple of $5$. By inspection, we can see that this is only possible if $y = 2^2k^5=4k^5$.
For $k=1$, $y=4$ and for $kge2$, $y>100$. So the only possible value of $y$ that fulfills the given conditions is $y=4$. However $y=4$ does not satisfy the equation $(i)$. So $y=4$ is also not admissible.
Hence there is no solution of the given equations such that $x, y in Z $ and $x,yle100$.
answered Jan 21 at 13:55
Faiq IrfanFaiq Irfan
814317
answered Jan 21 at 13:55
Faiq IrfanFaiq Irfan
814317
answered Jan 21 at 13:55
Faiq IrfanFaiq Irfan
814317
answered Jan 21 at 13:55
Faiq IrfanFaiq Irfan
814317
814317
$begingroup$
I tried to graph the two equations, I get two solutions (0,0) and (32,4) but I want an arithmetic explanation for it. Thank you.
$endgroup$
– Tanmay Gajapati
Jan 22 at 14:48
add a comment |
$begingroup$
I tried to graph the two equations, I get two solutions (0,0) and (32,4) but I want an arithmetic explanation for it. Thank you.
$endgroup$
– Tanmay Gajapati
Jan 22 at 14:48
$begingroup$
I tried to graph the two equations, I get two solutions (0,0) and (32,4) but I want an arithmetic explanation for it. Thank you.
$endgroup$
– Tanmay Gajapati
Jan 22 at 14:48
$begingroup$
I tried to graph the two equations, I get two solutions (0,0) and (32,4) but I want an arithmetic explanation for it. Thank you.
$endgroup$
– Tanmay Gajapati
Jan 22 at 14:48
add a comment |
1
$begingroup$
Well if I am not mistaken the second equation implies $y=frac x8$. Thus, you are left with an equation in $x$ alone, namely $$sqrt[5]{x}^3=frac{x^2}{64}-2$$
$endgroup$
– mrtaurho
Jan 21 at 13:42