Algebra divisibility proof
$begingroup$
How can i show that if $24a^2 +1 = b^2$ then $5$ divides $ab$??
I tried writing 24 as 25-1 but i did not get anything interesting
I have $25a^2=a^2+b^2-1$
But from there I don't know what to do
Any help would be appreciated
Thanks in advance!!
algebra-precalculus
asked Dec 27 '18 at 15:49
arsene steinarsene stein
1147
$endgroup$
add a comment |
$begingroup$
How can i show that if $24a^2 +1 = b^2$ then $5$ divides $ab$??
I tried writing 24 as 25-1 but i did not get anything interesting
I have $25a^2=a^2+b^2-1$
But from there I don't know what to do
Any help would be appreciated
Thanks in advance!!
algebra-precalculus
asked Dec 27 '18 at 15:49
arsene steinarsene stein
1147
$endgroup$
add a comment |
$begingroup$
How can i show that if $24a^2 +1 = b^2$ then $5$ divides $ab$??
I tried writing 24 as 25-1 but i did not get anything interesting
I have $25a^2=a^2+b^2-1$
But from there I don't know what to do
Any help would be appreciated
Thanks in advance!!
algebra-precalculus
asked Dec 27 '18 at 15:49
arsene steinarsene stein
1147
$endgroup$
How can i show that if $24a^2 +1 = b^2$ then $5$ divides $ab$??
I tried writing 24 as 25-1 but i did not get anything interesting
I have $25a^2=a^2+b^2-1$
But from there I don't know what to do
Any help would be appreciated
Thanks in advance!!
algebra-precalculus
algebra-precalculus
asked Dec 27 '18 at 15:49
arsene steinarsene stein
1147
asked Dec 27 '18 at 15:49
arsene steinarsene stein
1147
edited Dec 27 '18 at 16:24
arsene stein
asked Dec 27 '18 at 15:49
arsene steinarsene stein
1147
asked Dec 27 '18 at 15:49
arsene steinarsene stein
1147
asked Dec 27 '18 at 15:49
arsene steinarsene stein
1147
1147
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Ignoring the fact that the equation has no solution in integers, $a$ can not be divisible by $5$ because this would imply $5$ divides $5b^2-24a^2=1$, a contradiction. However $b$ is divisible by $5$. For $5b^2= 24a^2+1 = 25a^2-(a^2-1)$ implies $5$ divides $a^2-1$. Hence $a$ is either $1$ or $-1$ modulo $5$. But in either case you will find $24a^2+1$ is divisible by $25$. Thus $25$ divides $5b^2$ which implies $5$ divides $b^2$ which implies $5$ divides $b$, because $5$ is a prime.
answered Dec 27 '18 at 16:17
moutheticsmouthetics
52137
$endgroup$
$begingroup$
the question was edited about 7 minutes after your answer, it now says $24a^2 + 1 = b^2.$ Probably the most satisfactory thing for you is to add an answer to the revised question in this same answer box.
$endgroup$
– Will Jagy
Dec 27 '18 at 17:16
add a comment |
$begingroup$
Actually, $24a^2 + 1 = 5 b^2$ is not possible in integers.
480 factored 2^5 * 3 * 5
1. 1 20 -20 cycle length 2
2. -1 20 20 cycle length 2
3. 4 20 -5 cycle length 2
4. -4 20 5 cycle length 2
5. 3 18 -13 cycle length 4
6. -3 18 13 cycle length 4
7. 7 16 -8 cycle length 4
8. -7 16 8 cycle length 4
form class number is 8
answered Dec 27 '18 at 16:11
Will JagyWill Jagy
104k5103202
$endgroup$
$begingroup$
What of a=1 b=5??
$endgroup$
– arsene stein
Dec 27 '18 at 16:38
1
$begingroup$
@arsenestein it is quite rude to change a question after people have answered it. Note that the top lines of my answer refer explicitly to the original version of your question.
$endgroup$
– Will Jagy
Dec 27 '18 at 17:18
$begingroup$
Okay i see thanks
$endgroup$
– arsene stein
Dec 27 '18 at 17:20
add a comment |
$begingroup$
Here's a little fact: If $x$ is not divisible by $5$ then when you divide $x^2$ by $5$ the remainder is either $1$ or $4$. You can see this by noting that $x=5q+r$ for $r=1,2,3,4$ and squaring each case.
Now suppose that neither $a$ nor $b$ is divisible by $5$ and take your equation
$$25a^2 = a^2+b^2 -1.$$
The left side is a multiple of $5$, therefore so is the right side. Now try plugging in $1$ and $4$ for $a^2$ and $b^2$ to see if you can get a multiple of $5$ on the right. You can't. So you have a contradictions which shows that one of $a$ or $b$ has to be a multiple of $5$.
answered Dec 28 '18 at 10:21
B. GoddardB. Goddard
20.2k21543
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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$begingroup$
Ignoring the fact that the equation has no solution in integers, $a$ can not be divisible by $5$ because this would imply $5$ divides $5b^2-24a^2=1$, a contradiction. However $b$ is divisible by $5$. For $5b^2= 24a^2+1 = 25a^2-(a^2-1)$ implies $5$ divides $a^2-1$. Hence $a$ is either $1$ or $-1$ modulo $5$. But in either case you will find $24a^2+1$ is divisible by $25$. Thus $25$ divides $5b^2$ which implies $5$ divides $b^2$ which implies $5$ divides $b$, because $5$ is a prime.
answered Dec 27 '18 at 16:17
moutheticsmouthetics
52137
$endgroup$
$begingroup$
the question was edited about 7 minutes after your answer, it now says $24a^2 + 1 = b^2.$ Probably the most satisfactory thing for you is to add an answer to the revised question in this same answer box.
$endgroup$
– Will Jagy
Dec 27 '18 at 17:16
add a comment |
$begingroup$
Ignoring the fact that the equation has no solution in integers, $a$ can not be divisible by $5$ because this would imply $5$ divides $5b^2-24a^2=1$, a contradiction. However $b$ is divisible by $5$. For $5b^2= 24a^2+1 = 25a^2-(a^2-1)$ implies $5$ divides $a^2-1$. Hence $a$ is either $1$ or $-1$ modulo $5$. But in either case you will find $24a^2+1$ is divisible by $25$. Thus $25$ divides $5b^2$ which implies $5$ divides $b^2$ which implies $5$ divides $b$, because $5$ is a prime.
answered Dec 27 '18 at 16:17
moutheticsmouthetics
52137
$endgroup$
$begingroup$
the question was edited about 7 minutes after your answer, it now says $24a^2 + 1 = b^2.$ Probably the most satisfactory thing for you is to add an answer to the revised question in this same answer box.
$endgroup$
– Will Jagy
Dec 27 '18 at 17:16
add a comment |
$begingroup$
Ignoring the fact that the equation has no solution in integers, $a$ can not be divisible by $5$ because this would imply $5$ divides $5b^2-24a^2=1$, a contradiction. However $b$ is divisible by $5$. For $5b^2= 24a^2+1 = 25a^2-(a^2-1)$ implies $5$ divides $a^2-1$. Hence $a$ is either $1$ or $-1$ modulo $5$. But in either case you will find $24a^2+1$ is divisible by $25$. Thus $25$ divides $5b^2$ which implies $5$ divides $b^2$ which implies $5$ divides $b$, because $5$ is a prime.
answered Dec 27 '18 at 16:17
moutheticsmouthetics
52137
$endgroup$
Ignoring the fact that the equation has no solution in integers, $a$ can not be divisible by $5$ because this would imply $5$ divides $5b^2-24a^2=1$, a contradiction. However $b$ is divisible by $5$. For $5b^2= 24a^2+1 = 25a^2-(a^2-1)$ implies $5$ divides $a^2-1$. Hence $a$ is either $1$ or $-1$ modulo $5$. But in either case you will find $24a^2+1$ is divisible by $25$. Thus $25$ divides $5b^2$ which implies $5$ divides $b^2$ which implies $5$ divides $b$, because $5$ is a prime.
answered Dec 27 '18 at 16:17
moutheticsmouthetics
52137
answered Dec 27 '18 at 16:17
moutheticsmouthetics
52137
answered Dec 27 '18 at 16:17
moutheticsmouthetics
52137
answered Dec 27 '18 at 16:17
moutheticsmouthetics
52137
52137
$begingroup$
the question was edited about 7 minutes after your answer, it now says $24a^2 + 1 = b^2.$ Probably the most satisfactory thing for you is to add an answer to the revised question in this same answer box.
$endgroup$
– Will Jagy
Dec 27 '18 at 17:16
add a comment |
$begingroup$
the question was edited about 7 minutes after your answer, it now says $24a^2 + 1 = b^2.$ Probably the most satisfactory thing for you is to add an answer to the revised question in this same answer box.
$endgroup$
– Will Jagy
Dec 27 '18 at 17:16
$begingroup$
the question was edited about 7 minutes after your answer, it now says $24a^2 + 1 = b^2.$ Probably the most satisfactory thing for you is to add an answer to the revised question in this same answer box.
$endgroup$
– Will Jagy
Dec 27 '18 at 17:16
$begingroup$
the question was edited about 7 minutes after your answer, it now says $24a^2 + 1 = b^2.$ Probably the most satisfactory thing for you is to add an answer to the revised question in this same answer box.
$endgroup$
– Will Jagy
Dec 27 '18 at 17:16
add a comment |
$begingroup$
Actually, $24a^2 + 1 = 5 b^2$ is not possible in integers.
480 factored 2^5 * 3 * 5
1. 1 20 -20 cycle length 2
2. -1 20 20 cycle length 2
3. 4 20 -5 cycle length 2
4. -4 20 5 cycle length 2
5. 3 18 -13 cycle length 4
6. -3 18 13 cycle length 4
7. 7 16 -8 cycle length 4
8. -7 16 8 cycle length 4
form class number is 8
answered Dec 27 '18 at 16:11
Will JagyWill Jagy
104k5103202
$endgroup$
$begingroup$
What of a=1 b=5??
$endgroup$
– arsene stein
Dec 27 '18 at 16:38
1
$begingroup$
@arsenestein it is quite rude to change a question after people have answered it. Note that the top lines of my answer refer explicitly to the original version of your question.
$endgroup$
– Will Jagy
Dec 27 '18 at 17:18
$begingroup$
Okay i see thanks
$endgroup$
– arsene stein
Dec 27 '18 at 17:20
add a comment |
$begingroup$
Actually, $24a^2 + 1 = 5 b^2$ is not possible in integers.
480 factored 2^5 * 3 * 5
1. 1 20 -20 cycle length 2
2. -1 20 20 cycle length 2
3. 4 20 -5 cycle length 2
4. -4 20 5 cycle length 2
5. 3 18 -13 cycle length 4
6. -3 18 13 cycle length 4
7. 7 16 -8 cycle length 4
8. -7 16 8 cycle length 4
form class number is 8
answered Dec 27 '18 at 16:11
Will JagyWill Jagy
104k5103202
$endgroup$
$begingroup$
What of a=1 b=5??
$endgroup$
– arsene stein
Dec 27 '18 at 16:38
1
$begingroup$
@arsenestein it is quite rude to change a question after people have answered it. Note that the top lines of my answer refer explicitly to the original version of your question.
$endgroup$
– Will Jagy
Dec 27 '18 at 17:18
$begingroup$
Okay i see thanks
$endgroup$
– arsene stein
Dec 27 '18 at 17:20
add a comment |
$begingroup$
Actually, $24a^2 + 1 = 5 b^2$ is not possible in integers.
480 factored 2^5 * 3 * 5
1. 1 20 -20 cycle length 2
2. -1 20 20 cycle length 2
3. 4 20 -5 cycle length 2
4. -4 20 5 cycle length 2
5. 3 18 -13 cycle length 4
6. -3 18 13 cycle length 4
7. 7 16 -8 cycle length 4
8. -7 16 8 cycle length 4
form class number is 8
answered Dec 27 '18 at 16:11
Will JagyWill Jagy
104k5103202
$endgroup$
Actually, $24a^2 + 1 = 5 b^2$ is not possible in integers.
480 factored 2^5 * 3 * 5
1. 1 20 -20 cycle length 2
2. -1 20 20 cycle length 2
3. 4 20 -5 cycle length 2
4. -4 20 5 cycle length 2
5. 3 18 -13 cycle length 4
6. -3 18 13 cycle length 4
7. 7 16 -8 cycle length 4
8. -7 16 8 cycle length 4
form class number is 8
answered Dec 27 '18 at 16:11
Will JagyWill Jagy
104k5103202
answered Dec 27 '18 at 16:11
Will JagyWill Jagy
104k5103202
answered Dec 27 '18 at 16:11
Will JagyWill Jagy
104k5103202
answered Dec 27 '18 at 16:11
Will JagyWill Jagy
104k5103202
104k5103202
$begingroup$
What of a=1 b=5??
$endgroup$
– arsene stein
Dec 27 '18 at 16:38
1
$begingroup$
@arsenestein it is quite rude to change a question after people have answered it. Note that the top lines of my answer refer explicitly to the original version of your question.
$endgroup$
– Will Jagy
Dec 27 '18 at 17:18
$begingroup$
Okay i see thanks
$endgroup$
– arsene stein
Dec 27 '18 at 17:20
add a comment |
$begingroup$
What of a=1 b=5??
$endgroup$
– arsene stein
Dec 27 '18 at 16:38
1
$begingroup$
@arsenestein it is quite rude to change a question after people have answered it. Note that the top lines of my answer refer explicitly to the original version of your question.
$endgroup$
– Will Jagy
Dec 27 '18 at 17:18
$begingroup$
Okay i see thanks
$endgroup$
– arsene stein
Dec 27 '18 at 17:20
$begingroup$
What of a=1 b=5??
$endgroup$
– arsene stein
Dec 27 '18 at 16:38
$begingroup$
What of a=1 b=5??
$endgroup$
– arsene stein
Dec 27 '18 at 16:38
1
1
$begingroup$
@arsenestein it is quite rude to change a question after people have answered it. Note that the top lines of my answer refer explicitly to the original version of your question.
$endgroup$
– Will Jagy
Dec 27 '18 at 17:18
$begingroup$
@arsenestein it is quite rude to change a question after people have answered it. Note that the top lines of my answer refer explicitly to the original version of your question.
$endgroup$
– Will Jagy
Dec 27 '18 at 17:18
$begingroup$
Okay i see thanks
$endgroup$
– arsene stein
Dec 27 '18 at 17:20
$begingroup$
Okay i see thanks
$endgroup$
– arsene stein
Dec 27 '18 at 17:20
add a comment |
$begingroup$
Here's a little fact: If $x$ is not divisible by $5$ then when you divide $x^2$ by $5$ the remainder is either $1$ or $4$. You can see this by noting that $x=5q+r$ for $r=1,2,3,4$ and squaring each case.
Now suppose that neither $a$ nor $b$ is divisible by $5$ and take your equation
$$25a^2 = a^2+b^2 -1.$$
The left side is a multiple of $5$, therefore so is the right side. Now try plugging in $1$ and $4$ for $a^2$ and $b^2$ to see if you can get a multiple of $5$ on the right. You can't. So you have a contradictions which shows that one of $a$ or $b$ has to be a multiple of $5$.
answered Dec 28 '18 at 10:21
B. GoddardB. Goddard
20.2k21543
$endgroup$
add a comment |
$begingroup$
Here's a little fact: If $x$ is not divisible by $5$ then when you divide $x^2$ by $5$ the remainder is either $1$ or $4$. You can see this by noting that $x=5q+r$ for $r=1,2,3,4$ and squaring each case.
Now suppose that neither $a$ nor $b$ is divisible by $5$ and take your equation
$$25a^2 = a^2+b^2 -1.$$
The left side is a multiple of $5$, therefore so is the right side. Now try plugging in $1$ and $4$ for $a^2$ and $b^2$ to see if you can get a multiple of $5$ on the right. You can't. So you have a contradictions which shows that one of $a$ or $b$ has to be a multiple of $5$.
answered Dec 28 '18 at 10:21
B. GoddardB. Goddard
20.2k21543
$endgroup$
add a comment |
$begingroup$
Here's a little fact: If $x$ is not divisible by $5$ then when you divide $x^2$ by $5$ the remainder is either $1$ or $4$. You can see this by noting that $x=5q+r$ for $r=1,2,3,4$ and squaring each case.
Now suppose that neither $a$ nor $b$ is divisible by $5$ and take your equation
$$25a^2 = a^2+b^2 -1.$$
The left side is a multiple of $5$, therefore so is the right side. Now try plugging in $1$ and $4$ for $a^2$ and $b^2$ to see if you can get a multiple of $5$ on the right. You can't. So you have a contradictions which shows that one of $a$ or $b$ has to be a multiple of $5$.
answered Dec 28 '18 at 10:21
B. GoddardB. Goddard
20.2k21543
$endgroup$
Here's a little fact: If $x$ is not divisible by $5$ then when you divide $x^2$ by $5$ the remainder is either $1$ or $4$. You can see this by noting that $x=5q+r$ for $r=1,2,3,4$ and squaring each case.
Now suppose that neither $a$ nor $b$ is divisible by $5$ and take your equation
$$25a^2 = a^2+b^2 -1.$$
The left side is a multiple of $5$, therefore so is the right side. Now try plugging in $1$ and $4$ for $a^2$ and $b^2$ to see if you can get a multiple of $5$ on the right. You can't. So you have a contradictions which shows that one of $a$ or $b$ has to be a multiple of $5$.
answered Dec 28 '18 at 10:21
B. GoddardB. Goddard
20.2k21543
answered Dec 28 '18 at 10:21
B. GoddardB. Goddard
20.2k21543
answered Dec 28 '18 at 10:21
B. GoddardB. Goddard
20.2k21543
answered Dec 28 '18 at 10:21
B. GoddardB. Goddard
20.2k21543
20.2k21543
add a comment |
add a comment |
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