finding perfect squares solutions for the following case The Next CEO of Stack OverflowSubjects studied in number theoryCompute all the sets of 87248 into 10 partsIf $x^2+ax+b$ is an integer for every integer $x$ then comment on the coefficients $a$ and $b$ MCQNP-hardness of solving congruence equations in several variablesPairwise sums are perfect squares .When will the sequence $k mapsto A + Bk + k^2$ yield a perfect square?What are the integers of $textbf Q(sqrt 2 + i)$?Examples of $r^sqrt2$ as an irrational number, for real numbers $0<r<1$Calculating a square rootHow many perfect powers are there amoung the first 1000 positive integers
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finding perfect squares solutions for the following case
The Next CEO of Stack OverflowSubjects studied in number theoryCompute all the sets of 87248 into 10 partsIf $x^2+ax+b$ is an integer for every integer $x$ then comment on the coefficients $a$ and $b$ MCQNP-hardness of solving congruence equations in several variablesPairwise sums are perfect squares .When will the sequence $k mapsto A + Bk + k^2$ yield a perfect square?What are the integers of $textbf Q(sqrt 2 + i)$?Examples of $r^sqrt2$ as an irrational number, for real numbers $0<r<1$Calculating a square rootHow many perfect powers are there amoung the first 1000 positive integers
$begingroup$
I was working on a number theory problem and create a equation. I tried research on this, but tbh I don't even know what should I google for... Here's my cases.
$$n = sqrtN * frac1+sqrt4k^2+12$$
$$m = sqrtN * fracsqrt4k^2+1-12$$
Where N is a given integer, m, n are both unknown integers, k has a given range of [0, 10] and k is a real number.
My question is, What is the fastest way to find such k that create integers m and n?
Thanks in advance!
number-theory numerical-methods square-numbers
asked Mar 18 at 19:38
PetaGlzPetaGlz
53
$endgroup$
|
show 6 more comments
$begingroup$
I was working on a number theory problem and create a equation. I tried research on this, but tbh I don't even know what should I google for... Here's my cases.
$$n = sqrtN * frac1+sqrt4k^2+12$$
$$m = sqrtN * fracsqrt4k^2+1-12$$
Where N is a given integer, m, n are both unknown integers, k has a given range of [0, 10] and k is a real number.
My question is, What is the fastest way to find such k that create integers m and n?
Thanks in advance!
number-theory numerical-methods square-numbers
asked Mar 18 at 19:38
PetaGlzPetaGlz
53
$endgroup$
$begingroup$
Why would you expect the answer(s) to be unique? With $N=1$, for instance, then $k=2sqrt 3$ works as does $k=6sqrt 2$
$endgroup$
– lulu
Mar 18 at 19:42
$begingroup$
@lulu Ohh it doesn't have to be unique, should I change the 'n is an unknown integer' into 'unknown integers?
$endgroup$
– PetaGlz
Mar 18 at 19:46
$begingroup$
Your header refers to "unique" solutions, but maybe you meant something else?
$endgroup$
– lulu
Mar 18 at 19:47
$begingroup$
@lulu just changed that, thanks for that.
$endgroup$
– PetaGlz
Mar 18 at 19:48
$begingroup$
In any case: for a fixed $N$, I'd compute the expression with $k=frac 89$, and compute it for $k=9$. Then, for each integer between the two values you get you can easily find a solution $k$ that gives you that integer.
$endgroup$
– lulu
Mar 18 at 19:48
|
show 6 more comments
$begingroup$
I was working on a number theory problem and create a equation. I tried research on this, but tbh I don't even know what should I google for... Here's my cases.
$$n = sqrtN * frac1+sqrt4k^2+12$$
$$m = sqrtN * fracsqrt4k^2+1-12$$
Where N is a given integer, m, n are both unknown integers, k has a given range of [0, 10] and k is a real number.
My question is, What is the fastest way to find such k that create integers m and n?
Thanks in advance!
number-theory numerical-methods square-numbers
asked Mar 18 at 19:38
PetaGlzPetaGlz
53
$endgroup$
I was working on a number theory problem and create a equation. I tried research on this, but tbh I don't even know what should I google for... Here's my cases.
$$n = sqrtN * frac1+sqrt4k^2+12$$
$$m = sqrtN * fracsqrt4k^2+1-12$$
Where N is a given integer, m, n are both unknown integers, k has a given range of [0, 10] and k is a real number.
My question is, What is the fastest way to find such k that create integers m and n?
Thanks in advance!
number-theory numerical-methods square-numbers
number-theory numerical-methods square-numbers
asked Mar 18 at 19:38
PetaGlzPetaGlz
53
asked Mar 18 at 19:38
PetaGlzPetaGlz
53
edited Mar 18 at 20:26
PetaGlz
asked Mar 18 at 19:38
PetaGlzPetaGlz
53
asked Mar 18 at 19:38
PetaGlzPetaGlz
53
asked Mar 18 at 19:38
PetaGlzPetaGlz
53
53
$begingroup$
Why would you expect the answer(s) to be unique? With $N=1$, for instance, then $k=2sqrt 3$ works as does $k=6sqrt 2$
$endgroup$
– lulu
Mar 18 at 19:42
$begingroup$
@lulu Ohh it doesn't have to be unique, should I change the 'n is an unknown integer' into 'unknown integers?
$endgroup$
– PetaGlz
Mar 18 at 19:46
$begingroup$
Your header refers to "unique" solutions, but maybe you meant something else?
$endgroup$
– lulu
Mar 18 at 19:47
$begingroup$
@lulu just changed that, thanks for that.
$endgroup$
– PetaGlz
Mar 18 at 19:48
$begingroup$
In any case: for a fixed $N$, I'd compute the expression with $k=frac 89$, and compute it for $k=9$. Then, for each integer between the two values you get you can easily find a solution $k$ that gives you that integer.
$endgroup$
– lulu
Mar 18 at 19:48
|
show 6 more comments
$begingroup$
Why would you expect the answer(s) to be unique? With $N=1$, for instance, then $k=2sqrt 3$ works as does $k=6sqrt 2$
$endgroup$
– lulu
Mar 18 at 19:42
$begingroup$
@lulu Ohh it doesn't have to be unique, should I change the 'n is an unknown integer' into 'unknown integers?
$endgroup$
– PetaGlz
Mar 18 at 19:46
$begingroup$
Your header refers to "unique" solutions, but maybe you meant something else?
$endgroup$
– lulu
Mar 18 at 19:47
$begingroup$
@lulu just changed that, thanks for that.
$endgroup$
– PetaGlz
Mar 18 at 19:48
$begingroup$
In any case: for a fixed $N$, I'd compute the expression with $k=frac 89$, and compute it for $k=9$. Then, for each integer between the two values you get you can easily find a solution $k$ that gives you that integer.
$endgroup$
– lulu
Mar 18 at 19:48
$begingroup$
Why would you expect the answer(s) to be unique? With $N=1$, for instance, then $k=2sqrt 3$ works as does $k=6sqrt 2$
$endgroup$
– lulu
Mar 18 at 19:42
$begingroup$
Why would you expect the answer(s) to be unique? With $N=1$, for instance, then $k=2sqrt 3$ works as does $k=6sqrt 2$
$endgroup$
– lulu
Mar 18 at 19:42
$begingroup$
@lulu Ohh it doesn't have to be unique, should I change the 'n is an unknown integer' into 'unknown integers?
$endgroup$
– PetaGlz
Mar 18 at 19:46
$begingroup$
@lulu Ohh it doesn't have to be unique, should I change the 'n is an unknown integer' into 'unknown integers?
$endgroup$
– PetaGlz
Mar 18 at 19:46
$begingroup$
Your header refers to "unique" solutions, but maybe you meant something else?
$endgroup$
– lulu
Mar 18 at 19:47
$begingroup$
Your header refers to "unique" solutions, but maybe you meant something else?
$endgroup$
– lulu
Mar 18 at 19:47
$begingroup$
@lulu just changed that, thanks for that.
$endgroup$
– PetaGlz
Mar 18 at 19:48
$begingroup$
@lulu just changed that, thanks for that.
$endgroup$
– PetaGlz
Mar 18 at 19:48
$begingroup$
In any case: for a fixed $N$, I'd compute the expression with $k=frac 89$, and compute it for $k=9$. Then, for each integer between the two values you get you can easily find a solution $k$ that gives you that integer.
$endgroup$
– lulu
Mar 18 at 19:48
$begingroup$
In any case: for a fixed $N$, I'd compute the expression with $k=frac 89$, and compute it for $k=9$. Then, for each integer between the two values you get you can easily find a solution $k$ that gives you that integer.
$endgroup$
– lulu
Mar 18 at 19:48
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Given the range in $k$, you have $$1+frac sqrt33718 le frac 1+sqrt1+4k^22le 1+frac 5sqrt132$$
The left side is a little more than $2$ and the right a little more than $10$. The range of $n$ that is available is from $leftlceilsqrtleft(1+frac sqrt33718right)Nrightrceil$ to $leftlfloorsqrtleft(1+frac 5sqrt132right)Nrightrfloor$ inclusive. Choose your $n$ and solve the equation for $k$.
For your new problem with $m,n$, note that the radicands differ by $N$, so we can write $n^2-m^2=N=(n+m)(n-m)$. The two factors $n+m$ and $n-m$ have the same parity, so if $N$ is divisible by $2$ and not $4$ there is no solution. Otherwise, each way of factoring $N$ into two factors of the same parity give a solution to $n=sqrt aN+frac N2, m=sqrtaN-frac N2$. For each factorization, you can see if $a$ is in the range at the top of my post. If it is, $k$ will be in range, otherwise not.
answered Mar 18 at 20:09
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Thanks for your quick answer!
$endgroup$
– PetaGlz
Mar 18 at 20:28
$begingroup$
I just updated my problem since I realized that I wasn't described my problem well enough, sorry about that, could you check the update please?
$endgroup$
– PetaGlz
Mar 18 at 20:29
$begingroup$
Hey Ross! Thanks for the update!! the last formula involvong a is great, but I'm kinda wondering how did u get that. could you please explain that to me as well?
$endgroup$
– PetaGlz
Mar 18 at 20:53
$begingroup$
It is just the standard factoring of the difference of squares. You can multiply it out to see it works. For example, if $N=24=2^33$ we can have $n+m=12,n-m=2,n=7,m=5, n^2-m^2=49-25=24$ or $n+m=5,n-m=4,n=5,m=1,n^2-m^2=5^2-1^2=24$
$endgroup$
– Ross Millikan
Mar 18 at 20:57
$begingroup$
If you are doing number theory this factorization should be the first thing you think of when you see the difference of squares. It comes up all the time.
$endgroup$
– Ross Millikan
Mar 18 at 20:59
|
show 2 more comments
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Given the range in $k$, you have $$1+frac sqrt33718 le frac 1+sqrt1+4k^22le 1+frac 5sqrt132$$
The left side is a little more than $2$ and the right a little more than $10$. The range of $n$ that is available is from $leftlceilsqrtleft(1+frac sqrt33718right)Nrightrceil$ to $leftlfloorsqrtleft(1+frac 5sqrt132right)Nrightrfloor$ inclusive. Choose your $n$ and solve the equation for $k$.
For your new problem with $m,n$, note that the radicands differ by $N$, so we can write $n^2-m^2=N=(n+m)(n-m)$. The two factors $n+m$ and $n-m$ have the same parity, so if $N$ is divisible by $2$ and not $4$ there is no solution. Otherwise, each way of factoring $N$ into two factors of the same parity give a solution to $n=sqrt aN+frac N2, m=sqrtaN-frac N2$. For each factorization, you can see if $a$ is in the range at the top of my post. If it is, $k$ will be in range, otherwise not.
answered Mar 18 at 20:09
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Thanks for your quick answer!
$endgroup$
– PetaGlz
Mar 18 at 20:28
$begingroup$
I just updated my problem since I realized that I wasn't described my problem well enough, sorry about that, could you check the update please?
$endgroup$
– PetaGlz
Mar 18 at 20:29
$begingroup$
Hey Ross! Thanks for the update!! the last formula involvong a is great, but I'm kinda wondering how did u get that. could you please explain that to me as well?
$endgroup$
– PetaGlz
Mar 18 at 20:53
$begingroup$
It is just the standard factoring of the difference of squares. You can multiply it out to see it works. For example, if $N=24=2^33$ we can have $n+m=12,n-m=2,n=7,m=5, n^2-m^2=49-25=24$ or $n+m=5,n-m=4,n=5,m=1,n^2-m^2=5^2-1^2=24$
$endgroup$
– Ross Millikan
Mar 18 at 20:57
$begingroup$
If you are doing number theory this factorization should be the first thing you think of when you see the difference of squares. It comes up all the time.
$endgroup$
– Ross Millikan
Mar 18 at 20:59
|
show 2 more comments
$begingroup$
Given the range in $k$, you have $$1+frac sqrt33718 le frac 1+sqrt1+4k^22le 1+frac 5sqrt132$$
The left side is a little more than $2$ and the right a little more than $10$. The range of $n$ that is available is from $leftlceilsqrtleft(1+frac sqrt33718right)Nrightrceil$ to $leftlfloorsqrtleft(1+frac 5sqrt132right)Nrightrfloor$ inclusive. Choose your $n$ and solve the equation for $k$.
For your new problem with $m,n$, note that the radicands differ by $N$, so we can write $n^2-m^2=N=(n+m)(n-m)$. The two factors $n+m$ and $n-m$ have the same parity, so if $N$ is divisible by $2$ and not $4$ there is no solution. Otherwise, each way of factoring $N$ into two factors of the same parity give a solution to $n=sqrt aN+frac N2, m=sqrtaN-frac N2$. For each factorization, you can see if $a$ is in the range at the top of my post. If it is, $k$ will be in range, otherwise not.
answered Mar 18 at 20:09
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Thanks for your quick answer!
$endgroup$
– PetaGlz
Mar 18 at 20:28
$begingroup$
I just updated my problem since I realized that I wasn't described my problem well enough, sorry about that, could you check the update please?
$endgroup$
– PetaGlz
Mar 18 at 20:29
$begingroup$
Hey Ross! Thanks for the update!! the last formula involvong a is great, but I'm kinda wondering how did u get that. could you please explain that to me as well?
$endgroup$
– PetaGlz
Mar 18 at 20:53
$begingroup$
It is just the standard factoring of the difference of squares. You can multiply it out to see it works. For example, if $N=24=2^33$ we can have $n+m=12,n-m=2,n=7,m=5, n^2-m^2=49-25=24$ or $n+m=5,n-m=4,n=5,m=1,n^2-m^2=5^2-1^2=24$
$endgroup$
– Ross Millikan
Mar 18 at 20:57
$begingroup$
If you are doing number theory this factorization should be the first thing you think of when you see the difference of squares. It comes up all the time.
$endgroup$
– Ross Millikan
Mar 18 at 20:59
|
show 2 more comments
$begingroup$
Given the range in $k$, you have $$1+frac sqrt33718 le frac 1+sqrt1+4k^22le 1+frac 5sqrt132$$
The left side is a little more than $2$ and the right a little more than $10$. The range of $n$ that is available is from $leftlceilsqrtleft(1+frac sqrt33718right)Nrightrceil$ to $leftlfloorsqrtleft(1+frac 5sqrt132right)Nrightrfloor$ inclusive. Choose your $n$ and solve the equation for $k$.
For your new problem with $m,n$, note that the radicands differ by $N$, so we can write $n^2-m^2=N=(n+m)(n-m)$. The two factors $n+m$ and $n-m$ have the same parity, so if $N$ is divisible by $2$ and not $4$ there is no solution. Otherwise, each way of factoring $N$ into two factors of the same parity give a solution to $n=sqrt aN+frac N2, m=sqrtaN-frac N2$. For each factorization, you can see if $a$ is in the range at the top of my post. If it is, $k$ will be in range, otherwise not.
answered Mar 18 at 20:09
Ross MillikanRoss Millikan
300k24200375
$endgroup$
Given the range in $k$, you have $$1+frac sqrt33718 le frac 1+sqrt1+4k^22le 1+frac 5sqrt132$$
The left side is a little more than $2$ and the right a little more than $10$. The range of $n$ that is available is from $leftlceilsqrtleft(1+frac sqrt33718right)Nrightrceil$ to $leftlfloorsqrtleft(1+frac 5sqrt132right)Nrightrfloor$ inclusive. Choose your $n$ and solve the equation for $k$.
For your new problem with $m,n$, note that the radicands differ by $N$, so we can write $n^2-m^2=N=(n+m)(n-m)$. The two factors $n+m$ and $n-m$ have the same parity, so if $N$ is divisible by $2$ and not $4$ there is no solution. Otherwise, each way of factoring $N$ into two factors of the same parity give a solution to $n=sqrt aN+frac N2, m=sqrtaN-frac N2$. For each factorization, you can see if $a$ is in the range at the top of my post. If it is, $k$ will be in range, otherwise not.
answered Mar 18 at 20:09
Ross MillikanRoss Millikan
300k24200375
edited Mar 18 at 20:43
answered Mar 18 at 20:09
Ross MillikanRoss Millikan
300k24200375
answered Mar 18 at 20:09
Ross MillikanRoss Millikan
300k24200375
answered Mar 18 at 20:09
Ross MillikanRoss Millikan
300k24200375
300k24200375
$begingroup$
Thanks for your quick answer!
$endgroup$
– PetaGlz
Mar 18 at 20:28
$begingroup$
I just updated my problem since I realized that I wasn't described my problem well enough, sorry about that, could you check the update please?
$endgroup$
– PetaGlz
Mar 18 at 20:29
$begingroup$
Hey Ross! Thanks for the update!! the last formula involvong a is great, but I'm kinda wondering how did u get that. could you please explain that to me as well?
$endgroup$
– PetaGlz
Mar 18 at 20:53
$begingroup$
It is just the standard factoring of the difference of squares. You can multiply it out to see it works. For example, if $N=24=2^33$ we can have $n+m=12,n-m=2,n=7,m=5, n^2-m^2=49-25=24$ or $n+m=5,n-m=4,n=5,m=1,n^2-m^2=5^2-1^2=24$
$endgroup$
– Ross Millikan
Mar 18 at 20:57
$begingroup$
If you are doing number theory this factorization should be the first thing you think of when you see the difference of squares. It comes up all the time.
$endgroup$
– Ross Millikan
Mar 18 at 20:59
|
show 2 more comments
$begingroup$
Thanks for your quick answer!
$endgroup$
– PetaGlz
Mar 18 at 20:28
$begingroup$
I just updated my problem since I realized that I wasn't described my problem well enough, sorry about that, could you check the update please?
$endgroup$
– PetaGlz
Mar 18 at 20:29
$begingroup$
Hey Ross! Thanks for the update!! the last formula involvong a is great, but I'm kinda wondering how did u get that. could you please explain that to me as well?
$endgroup$
– PetaGlz
Mar 18 at 20:53
$begingroup$
It is just the standard factoring of the difference of squares. You can multiply it out to see it works. For example, if $N=24=2^33$ we can have $n+m=12,n-m=2,n=7,m=5, n^2-m^2=49-25=24$ or $n+m=5,n-m=4,n=5,m=1,n^2-m^2=5^2-1^2=24$
$endgroup$
– Ross Millikan
Mar 18 at 20:57
$begingroup$
If you are doing number theory this factorization should be the first thing you think of when you see the difference of squares. It comes up all the time.
$endgroup$
– Ross Millikan
Mar 18 at 20:59
$begingroup$
Thanks for your quick answer!
$endgroup$
– PetaGlz
Mar 18 at 20:28
$begingroup$
Thanks for your quick answer!
$endgroup$
– PetaGlz
Mar 18 at 20:28
$begingroup$
I just updated my problem since I realized that I wasn't described my problem well enough, sorry about that, could you check the update please?
$endgroup$
– PetaGlz
Mar 18 at 20:29
$begingroup$
I just updated my problem since I realized that I wasn't described my problem well enough, sorry about that, could you check the update please?
$endgroup$
– PetaGlz
Mar 18 at 20:29
$begingroup$
Hey Ross! Thanks for the update!! the last formula involvong a is great, but I'm kinda wondering how did u get that. could you please explain that to me as well?
$endgroup$
– PetaGlz
Mar 18 at 20:53
$begingroup$
Hey Ross! Thanks for the update!! the last formula involvong a is great, but I'm kinda wondering how did u get that. could you please explain that to me as well?
$endgroup$
– PetaGlz
Mar 18 at 20:53
$begingroup$
It is just the standard factoring of the difference of squares. You can multiply it out to see it works. For example, if $N=24=2^33$ we can have $n+m=12,n-m=2,n=7,m=5, n^2-m^2=49-25=24$ or $n+m=5,n-m=4,n=5,m=1,n^2-m^2=5^2-1^2=24$
$endgroup$
– Ross Millikan
Mar 18 at 20:57
$begingroup$
It is just the standard factoring of the difference of squares. You can multiply it out to see it works. For example, if $N=24=2^33$ we can have $n+m=12,n-m=2,n=7,m=5, n^2-m^2=49-25=24$ or $n+m=5,n-m=4,n=5,m=1,n^2-m^2=5^2-1^2=24$
$endgroup$
– Ross Millikan
Mar 18 at 20:57
$begingroup$
If you are doing number theory this factorization should be the first thing you think of when you see the difference of squares. It comes up all the time.
$endgroup$
– Ross Millikan
Mar 18 at 20:59
$begingroup$
If you are doing number theory this factorization should be the first thing you think of when you see the difference of squares. It comes up all the time.
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– Ross Millikan
Mar 18 at 20:59
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show 2 more comments
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Why would you expect the answer(s) to be unique? With $N=1$, for instance, then $k=2sqrt 3$ works as does $k=6sqrt 2$
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– lulu
Mar 18 at 19:42
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@lulu Ohh it doesn't have to be unique, should I change the 'n is an unknown integer' into 'unknown integers?
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– PetaGlz
Mar 18 at 19:46
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Your header refers to "unique" solutions, but maybe you meant something else?
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– lulu
Mar 18 at 19:47
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@lulu just changed that, thanks for that.
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– PetaGlz
Mar 18 at 19:48
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In any case: for a fixed $N$, I'd compute the expression with $k=frac 89$, and compute it for $k=9$. Then, for each integer between the two values you get you can easily find a solution $k$ that gives you that integer.
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– lulu
Mar 18 at 19:48