Solving for solutions to $left(a+bright)^2=a.b$Solve $frac{b}{a}=a.b$ in decimalSolve $x^4 - 2x^3 + x = y^4 +...
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Solving for solutions to $left(a+bright)^2=a.b$
Solve $frac{b}{a}=a.b$ in decimalSolve $x^4 - 2x^3 + x = y^4 + 3y^2 + y wedge (x,y) in mathbb{Z}^2$What is the explicit formula (solution) to this recursively defined binary matrix?integer solutions for $left( a+sqrt{b} right) ^ n=p+qsqrt{b}$?An alternative proof for Bertrand's Postulate when $n ge 36$$G(n) = 1 + sumlimits_{ substack{d|n\d != 1\dneq n}}G(n /d)$, how to solve for $G(n)=n$.Why is it so hard to find numbers with some palindromic properties?How to prove $logleft(frac e{2t}right) approx logleft(frac1tright)$ for $0 < t < frac 12$?Patterns in solutions to simultaneous palindromes in two number basesSolving for $x$ in $y=Ncdotleft(frac{10}{x}right)^{-2.6}$
$begingroup$
I ran across this stackoverflow question regarding finding solutions to
$$left(a+bright)^2=a.b$$
for $a,binBbb{N}$ where $a.b$ denotes concatenation of $a$ and $b$. Since concatenation isn't really a "true" mathematical operation, I'll write
$$left(a+bright)^2=atimes10^c+b$$
where $c=lfloor{log_{10}b}rfloor+1$. The post there also has an implicit stipulation that neither $a$ nor $b$ may contain leading zeros, but that does not matter for the sake of this question- I can filter those solutions later. I wish to gain some mathematical insight into the pattern of these solutions, so that maybe I can write a more efficient solver.
The naive method of searching for solutions by iterating over pairs of $a$ and $b$ takes $Oleft(n^2right)$ time. By instead iterating over all squares $i^2$, and decomposing each into $lfloor2log_{10}irfloor$ separate $a,b$ pairs, we can search for solutions in $Oleft(nlog(n)right)$ time. I've since realized that because $a,b<i$ and $atimes10^c+b=i^2$, that $a$ and $b$ must both be within $+0/-2$ orders of magnitude of $i$, and so there is at most three possibly valid $a,b$ pairs per $i^2$. This brings the time complexity down to $Oleft(nright)$, but I still feel like there might be other mathematical insights that could allow me to prune other invalid $a,b$, whether they result in an asymptotic speedup or not.
Is there a closed-form solution to this problem, or any particular pattern to the $a$ and $b$ forming valid solutions?
algebra-precalculus elementary-number-theory
asked Mar 4 at 2:17
Dillon DavisDillon Davis
1315
New contributor
Dillon Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 3 more comments
$begingroup$
I ran across this stackoverflow question regarding finding solutions to
$$left(a+bright)^2=a.b$$
for $a,binBbb{N}$ where $a.b$ denotes concatenation of $a$ and $b$. Since concatenation isn't really a "true" mathematical operation, I'll write
$$left(a+bright)^2=atimes10^c+b$$
where $c=lfloor{log_{10}b}rfloor+1$. The post there also has an implicit stipulation that neither $a$ nor $b$ may contain leading zeros, but that does not matter for the sake of this question- I can filter those solutions later. I wish to gain some mathematical insight into the pattern of these solutions, so that maybe I can write a more efficient solver.
The naive method of searching for solutions by iterating over pairs of $a$ and $b$ takes $Oleft(n^2right)$ time. By instead iterating over all squares $i^2$, and decomposing each into $lfloor2log_{10}irfloor$ separate $a,b$ pairs, we can search for solutions in $Oleft(nlog(n)right)$ time. I've since realized that because $a,b<i$ and $atimes10^c+b=i^2$, that $a$ and $b$ must both be within $+0/-2$ orders of magnitude of $i$, and so there is at most three possibly valid $a,b$ pairs per $i^2$. This brings the time complexity down to $Oleft(nright)$, but I still feel like there might be other mathematical insights that could allow me to prune other invalid $a,b$, whether they result in an asymptotic speedup or not.
Is there a closed-form solution to this problem, or any particular pattern to the $a$ and $b$ forming valid solutions?
algebra-precalculus elementary-number-theory
asked Mar 4 at 2:17
Dillon DavisDillon Davis
1315
New contributor
Dillon Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Reducing modulo $a$ gives $b^2 equiv b pmod a$, and hence $b^2 equiv b mod m$ if $m mid a$. For generic $a$ this leads to strong restrictions on $b$. For prime $a = p$ (or prime factors $p$ of $a$) this implies $b = 0, 1 pmod p$.
$endgroup$
– Travis
Mar 4 at 2:26
2
$begingroup$
Your formula for concatenation isn't quite right. It's problematic when $b$ is a power of $10$. E.g., if $a=1$ and $b=10$, then $c=lceillog_{10}10rceil=1$ and $atimes10^c+b=20$, not $110$.
$endgroup$
– Barry Cipra
Mar 4 at 2:53
2
$begingroup$
One can just as well prune cases by reducing modulo $b$, which gives $a^2 = 10^c a pmod b$. If $2 mid b$, we get immediately that $2 mid a$, likewise for $5$. Similarly, if $3 mid b$, then we conclude that either $3 mid a$ or $a equiv (-1)^c pmod 3$.
$endgroup$
– Travis
Mar 4 at 3:22
2
$begingroup$
One simple pruning: $a$ must be even, since $(a+b)^2equiv a+b$ mod $2$. So you might as well rewrite the equation to solve as $(2a+b)^2=2atimes10^c+b$. Also, $b$ cannot be congruent to $2$ or $3$ mod $5$, since $(a+b)^2$ is always $0$, $1$, or $4$.
$endgroup$
– Barry Cipra
Mar 4 at 13:01
2
$begingroup$
One other simple pruning: As soon as $b$ has two or more digits, it must be congruent to $0$, $1$, or $4$ mod $8$, since those are the only residues for $(a+b)^2$ (and $a$, as noted in the previous column, is even, so $atimes10^cequiv0$ mod $8$ if $cge2$).
$endgroup$
– Barry Cipra
Mar 4 at 13:12
|
show 3 more comments
$begingroup$
I ran across this stackoverflow question regarding finding solutions to
$$left(a+bright)^2=a.b$$
for $a,binBbb{N}$ where $a.b$ denotes concatenation of $a$ and $b$. Since concatenation isn't really a "true" mathematical operation, I'll write
$$left(a+bright)^2=atimes10^c+b$$
where $c=lfloor{log_{10}b}rfloor+1$. The post there also has an implicit stipulation that neither $a$ nor $b$ may contain leading zeros, but that does not matter for the sake of this question- I can filter those solutions later. I wish to gain some mathematical insight into the pattern of these solutions, so that maybe I can write a more efficient solver.
The naive method of searching for solutions by iterating over pairs of $a$ and $b$ takes $Oleft(n^2right)$ time. By instead iterating over all squares $i^2$, and decomposing each into $lfloor2log_{10}irfloor$ separate $a,b$ pairs, we can search for solutions in $Oleft(nlog(n)right)$ time. I've since realized that because $a,b<i$ and $atimes10^c+b=i^2$, that $a$ and $b$ must both be within $+0/-2$ orders of magnitude of $i$, and so there is at most three possibly valid $a,b$ pairs per $i^2$. This brings the time complexity down to $Oleft(nright)$, but I still feel like there might be other mathematical insights that could allow me to prune other invalid $a,b$, whether they result in an asymptotic speedup or not.
Is there a closed-form solution to this problem, or any particular pattern to the $a$ and $b$ forming valid solutions?
algebra-precalculus elementary-number-theory
asked Mar 4 at 2:17
Dillon DavisDillon Davis
1315
New contributor
Dillon Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I ran across this stackoverflow question regarding finding solutions to
$$left(a+bright)^2=a.b$$
for $a,binBbb{N}$ where $a.b$ denotes concatenation of $a$ and $b$. Since concatenation isn't really a "true" mathematical operation, I'll write
$$left(a+bright)^2=atimes10^c+b$$
where $c=lfloor{log_{10}b}rfloor+1$. The post there also has an implicit stipulation that neither $a$ nor $b$ may contain leading zeros, but that does not matter for the sake of this question- I can filter those solutions later. I wish to gain some mathematical insight into the pattern of these solutions, so that maybe I can write a more efficient solver.
The naive method of searching for solutions by iterating over pairs of $a$ and $b$ takes $Oleft(n^2right)$ time. By instead iterating over all squares $i^2$, and decomposing each into $lfloor2log_{10}irfloor$ separate $a,b$ pairs, we can search for solutions in $Oleft(nlog(n)right)$ time. I've since realized that because $a,b<i$ and $atimes10^c+b=i^2$, that $a$ and $b$ must both be within $+0/-2$ orders of magnitude of $i$, and so there is at most three possibly valid $a,b$ pairs per $i^2$. This brings the time complexity down to $Oleft(nright)$, but I still feel like there might be other mathematical insights that could allow me to prune other invalid $a,b$, whether they result in an asymptotic speedup or not.
Is there a closed-form solution to this problem, or any particular pattern to the $a$ and $b$ forming valid solutions?
algebra-precalculus elementary-number-theory
algebra-precalculus elementary-number-theory
asked Mar 4 at 2:17
Dillon DavisDillon Davis
1315
New contributor
Dillon Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 4 at 2:17
Dillon DavisDillon Davis
1315
New contributor
Dillon Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 4 at 2:56
Dillon Davis
asked Mar 4 at 2:17
Dillon DavisDillon Davis
1315
New contributor
Dillon Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 4 at 2:17
Dillon DavisDillon Davis
1315
asked Mar 4 at 2:17
Dillon DavisDillon Davis
1315
1315
New contributor
Dillon Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dillon Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dillon Davis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Reducing modulo $a$ gives $b^2 equiv b pmod a$, and hence $b^2 equiv b mod m$ if $m mid a$. For generic $a$ this leads to strong restrictions on $b$. For prime $a = p$ (or prime factors $p$ of $a$) this implies $b = 0, 1 pmod p$.
$endgroup$
– Travis
Mar 4 at 2:26
2
$begingroup$
Your formula for concatenation isn't quite right. It's problematic when $b$ is a power of $10$. E.g., if $a=1$ and $b=10$, then $c=lceillog_{10}10rceil=1$ and $atimes10^c+b=20$, not $110$.
$endgroup$
– Barry Cipra
Mar 4 at 2:53
2
$begingroup$
One can just as well prune cases by reducing modulo $b$, which gives $a^2 = 10^c a pmod b$. If $2 mid b$, we get immediately that $2 mid a$, likewise for $5$. Similarly, if $3 mid b$, then we conclude that either $3 mid a$ or $a equiv (-1)^c pmod 3$.
$endgroup$
– Travis
Mar 4 at 3:22
2
$begingroup$
One simple pruning: $a$ must be even, since $(a+b)^2equiv a+b$ mod $2$. So you might as well rewrite the equation to solve as $(2a+b)^2=2atimes10^c+b$. Also, $b$ cannot be congruent to $2$ or $3$ mod $5$, since $(a+b)^2$ is always $0$, $1$, or $4$.
$endgroup$
– Barry Cipra
Mar 4 at 13:01
2
$begingroup$
One other simple pruning: As soon as $b$ has two or more digits, it must be congruent to $0$, $1$, or $4$ mod $8$, since those are the only residues for $(a+b)^2$ (and $a$, as noted in the previous column, is even, so $atimes10^cequiv0$ mod $8$ if $cge2$).
$endgroup$
– Barry Cipra
Mar 4 at 13:12
|
show 3 more comments
2
$begingroup$
Reducing modulo $a$ gives $b^2 equiv b pmod a$, and hence $b^2 equiv b mod m$ if $m mid a$. For generic $a$ this leads to strong restrictions on $b$. For prime $a = p$ (or prime factors $p$ of $a$) this implies $b = 0, 1 pmod p$.
$endgroup$
– Travis
Mar 4 at 2:26
2
$begingroup$
Your formula for concatenation isn't quite right. It's problematic when $b$ is a power of $10$. E.g., if $a=1$ and $b=10$, then $c=lceillog_{10}10rceil=1$ and $atimes10^c+b=20$, not $110$.
$endgroup$
– Barry Cipra
Mar 4 at 2:53
2
$begingroup$
One can just as well prune cases by reducing modulo $b$, which gives $a^2 = 10^c a pmod b$. If $2 mid b$, we get immediately that $2 mid a$, likewise for $5$. Similarly, if $3 mid b$, then we conclude that either $3 mid a$ or $a equiv (-1)^c pmod 3$.
$endgroup$
– Travis
Mar 4 at 3:22
2
$begingroup$
One simple pruning: $a$ must be even, since $(a+b)^2equiv a+b$ mod $2$. So you might as well rewrite the equation to solve as $(2a+b)^2=2atimes10^c+b$. Also, $b$ cannot be congruent to $2$ or $3$ mod $5$, since $(a+b)^2$ is always $0$, $1$, or $4$.
$endgroup$
– Barry Cipra
Mar 4 at 13:01
2
$begingroup$
One other simple pruning: As soon as $b$ has two or more digits, it must be congruent to $0$, $1$, or $4$ mod $8$, since those are the only residues for $(a+b)^2$ (and $a$, as noted in the previous column, is even, so $atimes10^cequiv0$ mod $8$ if $cge2$).
$endgroup$
– Barry Cipra
Mar 4 at 13:12
2
2
$begingroup$
Reducing modulo $a$ gives $b^2 equiv b pmod a$, and hence $b^2 equiv b mod m$ if $m mid a$. For generic $a$ this leads to strong restrictions on $b$. For prime $a = p$ (or prime factors $p$ of $a$) this implies $b = 0, 1 pmod p$.
$endgroup$
– Travis
Mar 4 at 2:26
$begingroup$
Reducing modulo $a$ gives $b^2 equiv b pmod a$, and hence $b^2 equiv b mod m$ if $m mid a$. For generic $a$ this leads to strong restrictions on $b$. For prime $a = p$ (or prime factors $p$ of $a$) this implies $b = 0, 1 pmod p$.
$endgroup$
– Travis
Mar 4 at 2:26
2
2
$begingroup$
Your formula for concatenation isn't quite right. It's problematic when $b$ is a power of $10$. E.g., if $a=1$ and $b=10$, then $c=lceillog_{10}10rceil=1$ and $atimes10^c+b=20$, not $110$.
$endgroup$
– Barry Cipra
Mar 4 at 2:53
$begingroup$
Your formula for concatenation isn't quite right. It's problematic when $b$ is a power of $10$. E.g., if $a=1$ and $b=10$, then $c=lceillog_{10}10rceil=1$ and $atimes10^c+b=20$, not $110$.
$endgroup$
– Barry Cipra
Mar 4 at 2:53
2
2
$begingroup$
One can just as well prune cases by reducing modulo $b$, which gives $a^2 = 10^c a pmod b$. If $2 mid b$, we get immediately that $2 mid a$, likewise for $5$. Similarly, if $3 mid b$, then we conclude that either $3 mid a$ or $a equiv (-1)^c pmod 3$.
$endgroup$
– Travis
Mar 4 at 3:22
$begingroup$
One can just as well prune cases by reducing modulo $b$, which gives $a^2 = 10^c a pmod b$. If $2 mid b$, we get immediately that $2 mid a$, likewise for $5$. Similarly, if $3 mid b$, then we conclude that either $3 mid a$ or $a equiv (-1)^c pmod 3$.
$endgroup$
– Travis
Mar 4 at 3:22
2
2
$begingroup$
One simple pruning: $a$ must be even, since $(a+b)^2equiv a+b$ mod $2$. So you might as well rewrite the equation to solve as $(2a+b)^2=2atimes10^c+b$. Also, $b$ cannot be congruent to $2$ or $3$ mod $5$, since $(a+b)^2$ is always $0$, $1$, or $4$.
$endgroup$
– Barry Cipra
Mar 4 at 13:01
$begingroup$
One simple pruning: $a$ must be even, since $(a+b)^2equiv a+b$ mod $2$. So you might as well rewrite the equation to solve as $(2a+b)^2=2atimes10^c+b$. Also, $b$ cannot be congruent to $2$ or $3$ mod $5$, since $(a+b)^2$ is always $0$, $1$, or $4$.
$endgroup$
– Barry Cipra
Mar 4 at 13:01
2
2
$begingroup$
One other simple pruning: As soon as $b$ has two or more digits, it must be congruent to $0$, $1$, or $4$ mod $8$, since those are the only residues for $(a+b)^2$ (and $a$, as noted in the previous column, is even, so $atimes10^cequiv0$ mod $8$ if $cge2$).
$endgroup$
– Barry Cipra
Mar 4 at 13:12
$begingroup$
One other simple pruning: As soon as $b$ has two or more digits, it must be congruent to $0$, $1$, or $4$ mod $8$, since those are the only residues for $(a+b)^2$ (and $a$, as noted in the previous column, is even, so $atimes10^cequiv0$ mod $8$ if $cge2$).
$endgroup$
– Barry Cipra
Mar 4 at 13:12
|
show 3 more comments
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Dillon Davis is a new contributor. Be nice, and check out our Code of Conduct.
Dillon Davis is a new contributor. Be nice, and check out our Code of Conduct.
Dillon Davis is a new contributor. Be nice, and check out our Code of Conduct.
Dillon Davis is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
Reducing modulo $a$ gives $b^2 equiv b pmod a$, and hence $b^2 equiv b mod m$ if $m mid a$. For generic $a$ this leads to strong restrictions on $b$. For prime $a = p$ (or prime factors $p$ of $a$) this implies $b = 0, 1 pmod p$.
$endgroup$
– Travis
Mar 4 at 2:26
2
$begingroup$
Your formula for concatenation isn't quite right. It's problematic when $b$ is a power of $10$. E.g., if $a=1$ and $b=10$, then $c=lceillog_{10}10rceil=1$ and $atimes10^c+b=20$, not $110$.
$endgroup$
– Barry Cipra
Mar 4 at 2:53
2
$begingroup$
One can just as well prune cases by reducing modulo $b$, which gives $a^2 = 10^c a pmod b$. If $2 mid b$, we get immediately that $2 mid a$, likewise for $5$. Similarly, if $3 mid b$, then we conclude that either $3 mid a$ or $a equiv (-1)^c pmod 3$.
$endgroup$
– Travis
Mar 4 at 3:22
2
$begingroup$
One simple pruning: $a$ must be even, since $(a+b)^2equiv a+b$ mod $2$. So you might as well rewrite the equation to solve as $(2a+b)^2=2atimes10^c+b$. Also, $b$ cannot be congruent to $2$ or $3$ mod $5$, since $(a+b)^2$ is always $0$, $1$, or $4$.
$endgroup$
– Barry Cipra
Mar 4 at 13:01
2
$begingroup$
One other simple pruning: As soon as $b$ has two or more digits, it must be congruent to $0$, $1$, or $4$ mod $8$, since those are the only residues for $(a+b)^2$ (and $a$, as noted in the previous column, is even, so $atimes10^cequiv0$ mod $8$ if $cge2$).
$endgroup$
– Barry Cipra
Mar 4 at 13:12