Finding elements such that none add to a perfect square
Bob asks us to find an infinite set $S$ of positive integers such that the sum of any finite number of distinct elements of S is not a perfect square.
Can Bob's request be fulfilled?
I can find some finite sets, and the sequence A133662 (OEIS) seems to work but I don't know if that sequence is infinite or not.
Maybe if we picked lots of elements with a common property?
sequences-and-series number-theory square-numbers sumset
asked Dec 23 '18 at 0:03
user627514
423
add a comment |
Bob asks us to find an infinite set $S$ of positive integers such that the sum of any finite number of distinct elements of S is not a perfect square.
Can Bob's request be fulfilled?
I can find some finite sets, and the sequence A133662 (OEIS) seems to work but I don't know if that sequence is infinite or not.
Maybe if we picked lots of elements with a common property?
sequences-and-series number-theory square-numbers sumset
asked Dec 23 '18 at 0:03
user627514
423
add a comment |
Bob asks us to find an infinite set $S$ of positive integers such that the sum of any finite number of distinct elements of S is not a perfect square.
Can Bob's request be fulfilled?
I can find some finite sets, and the sequence A133662 (OEIS) seems to work but I don't know if that sequence is infinite or not.
Maybe if we picked lots of elements with a common property?
sequences-and-series number-theory square-numbers sumset
asked Dec 23 '18 at 0:03
user627514
423
Bob asks us to find an infinite set $S$ of positive integers such that the sum of any finite number of distinct elements of S is not a perfect square.
Can Bob's request be fulfilled?
I can find some finite sets, and the sequence A133662 (OEIS) seems to work but I don't know if that sequence is infinite or not.
Maybe if we picked lots of elements with a common property?
sequences-and-series number-theory square-numbers sumset
sequences-and-series number-theory square-numbers sumset
asked Dec 23 '18 at 0:03
user627514
423
asked Dec 23 '18 at 0:03
user627514
423
asked Dec 23 '18 at 0:03
user627514
423
asked Dec 23 '18 at 0:03
user627514
423
asked Dec 23 '18 at 0:03
user627514
423
423
add a comment |
add a comment |
2 Answers
2
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Pick a prime $p$ and consider $S_p={p,p^3,p^5,cdots}$.
Let $v_p(n)$ denote the maximal power of $p$ which divides $n$, so $v_3(18)=2$ for example.
Any finite sum of elements in $S_p$ has the form $m=p^{2a_1+1}+cdots p^{2a_k+1}$ with $a_1<a_2<cdots <a_k$. But we then have $$v_pleft(mright)=v_pleft(p^{2a_1+1}+cdots p^{2a_k+1}right)=2a_1+1$$ which is odd, so $m$ can not be a perfect square.
answered Dec 23 '18 at 0:08
lulu
39.2k24677
$v_p(m)=2a_k+1$ since $a_1<dots<a_k$. But very nice! I have never seen $v_p$ method used before, do you have some literature about it?
– user627514
Dec 23 '18 at 0:12
1
The notation is fairly standard, at least in the context of valuation theory. But the concept is completely standard. If you take the unique factorization of $m$ then $v_p(m)$ is just the exponent of $p$ in that expression.
– lulu
Dec 23 '18 at 0:14
add a comment |
Pick any sequence $a_1,a_2,ldots$ which grows fast enough so that $a_1+a_2+ldots a_{n-1}$ is smaller than the distance from $a_n$ up to the next perfect square (and $a_n$ should itself never be a perfect square, of course). That way, any finite sum of terms will have a largest term, and the rest of the terms in the sum won't be enough to reach the next square from there.
So, for instance, let $a_1=2$. Then we can let $a_2=5$, because the next square from $5$ is $9$, and $2+5<9$.
From here, $2+5=7$, so the next candidate for $a_3$ is $17$, as the next square up from $17$ is $25$, and you can't reach $25$ from $17$ by adding $2$ and / or $5$.
We now have $2+5+17=24$, so the next candidate is $170$, as the next square from there is $196$, and $2,5,17$ aren't large enough to reach that high.
This can continue indefinitely. The gaps between consecutive squares grow, so you can always find a large enough gap to put the next element of your sequence.
answered Dec 23 '18 at 0:17
Arthur
111k7105186
2
One can, of course, be more compact. For instance, take $a_1=2,a_2=3,a_3=5,a_4=10,ldots$, where each term is the lowest possible number which doesn't allow any sum of terms to be a perfect square. But it quickly becomes tricky to keep track of everything to correctly find the next number in the sequence (and one even has to do an argument similar to the one above to prove that there is always a next term), so I like the approach in the answer better for its simplicity.
– Arthur
Dec 23 '18 at 0:31
add a comment |
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2 Answers
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2 Answers
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Pick a prime $p$ and consider $S_p={p,p^3,p^5,cdots}$.
Let $v_p(n)$ denote the maximal power of $p$ which divides $n$, so $v_3(18)=2$ for example.
Any finite sum of elements in $S_p$ has the form $m=p^{2a_1+1}+cdots p^{2a_k+1}$ with $a_1<a_2<cdots <a_k$. But we then have $$v_pleft(mright)=v_pleft(p^{2a_1+1}+cdots p^{2a_k+1}right)=2a_1+1$$ which is odd, so $m$ can not be a perfect square.
answered Dec 23 '18 at 0:08
lulu
39.2k24677
$v_p(m)=2a_k+1$ since $a_1<dots<a_k$. But very nice! I have never seen $v_p$ method used before, do you have some literature about it?
– user627514
Dec 23 '18 at 0:12
1
The notation is fairly standard, at least in the context of valuation theory. But the concept is completely standard. If you take the unique factorization of $m$ then $v_p(m)$ is just the exponent of $p$ in that expression.
– lulu
Dec 23 '18 at 0:14
add a comment |
Pick a prime $p$ and consider $S_p={p,p^3,p^5,cdots}$.
Let $v_p(n)$ denote the maximal power of $p$ which divides $n$, so $v_3(18)=2$ for example.
Any finite sum of elements in $S_p$ has the form $m=p^{2a_1+1}+cdots p^{2a_k+1}$ with $a_1<a_2<cdots <a_k$. But we then have $$v_pleft(mright)=v_pleft(p^{2a_1+1}+cdots p^{2a_k+1}right)=2a_1+1$$ which is odd, so $m$ can not be a perfect square.
answered Dec 23 '18 at 0:08
lulu
39.2k24677
$v_p(m)=2a_k+1$ since $a_1<dots<a_k$. But very nice! I have never seen $v_p$ method used before, do you have some literature about it?
– user627514
Dec 23 '18 at 0:12
1
The notation is fairly standard, at least in the context of valuation theory. But the concept is completely standard. If you take the unique factorization of $m$ then $v_p(m)$ is just the exponent of $p$ in that expression.
– lulu
Dec 23 '18 at 0:14
add a comment |
Pick a prime $p$ and consider $S_p={p,p^3,p^5,cdots}$.
Let $v_p(n)$ denote the maximal power of $p$ which divides $n$, so $v_3(18)=2$ for example.
Any finite sum of elements in $S_p$ has the form $m=p^{2a_1+1}+cdots p^{2a_k+1}$ with $a_1<a_2<cdots <a_k$. But we then have $$v_pleft(mright)=v_pleft(p^{2a_1+1}+cdots p^{2a_k+1}right)=2a_1+1$$ which is odd, so $m$ can not be a perfect square.
answered Dec 23 '18 at 0:08
lulu
39.2k24677
Pick a prime $p$ and consider $S_p={p,p^3,p^5,cdots}$.
Let $v_p(n)$ denote the maximal power of $p$ which divides $n$, so $v_3(18)=2$ for example.
Any finite sum of elements in $S_p$ has the form $m=p^{2a_1+1}+cdots p^{2a_k+1}$ with $a_1<a_2<cdots <a_k$. But we then have $$v_pleft(mright)=v_pleft(p^{2a_1+1}+cdots p^{2a_k+1}right)=2a_1+1$$ which is odd, so $m$ can not be a perfect square.
answered Dec 23 '18 at 0:08
lulu
39.2k24677
edited Dec 23 '18 at 0:14
answered Dec 23 '18 at 0:08
lulu
39.2k24677
answered Dec 23 '18 at 0:08
lulu
39.2k24677
answered Dec 23 '18 at 0:08
lulu
39.2k24677
39.2k24677
$v_p(m)=2a_k+1$ since $a_1<dots<a_k$. But very nice! I have never seen $v_p$ method used before, do you have some literature about it?
– user627514
Dec 23 '18 at 0:12
1
The notation is fairly standard, at least in the context of valuation theory. But the concept is completely standard. If you take the unique factorization of $m$ then $v_p(m)$ is just the exponent of $p$ in that expression.
– lulu
Dec 23 '18 at 0:14
add a comment |
$v_p(m)=2a_k+1$ since $a_1<dots<a_k$. But very nice! I have never seen $v_p$ method used before, do you have some literature about it?
– user627514
Dec 23 '18 at 0:12
1
The notation is fairly standard, at least in the context of valuation theory. But the concept is completely standard. If you take the unique factorization of $m$ then $v_p(m)$ is just the exponent of $p$ in that expression.
– lulu
Dec 23 '18 at 0:14
$v_p(m)=2a_k+1$ since $a_1<dots<a_k$. But very nice! I have never seen $v_p$ method used before, do you have some literature about it?
– user627514
Dec 23 '18 at 0:12
$v_p(m)=2a_k+1$ since $a_1<dots<a_k$. But very nice! I have never seen $v_p$ method used before, do you have some literature about it?
– user627514
Dec 23 '18 at 0:12
1
1
The notation is fairly standard, at least in the context of valuation theory. But the concept is completely standard. If you take the unique factorization of $m$ then $v_p(m)$ is just the exponent of $p$ in that expression.
– lulu
Dec 23 '18 at 0:14
The notation is fairly standard, at least in the context of valuation theory. But the concept is completely standard. If you take the unique factorization of $m$ then $v_p(m)$ is just the exponent of $p$ in that expression.
– lulu
Dec 23 '18 at 0:14
add a comment |
Pick any sequence $a_1,a_2,ldots$ which grows fast enough so that $a_1+a_2+ldots a_{n-1}$ is smaller than the distance from $a_n$ up to the next perfect square (and $a_n$ should itself never be a perfect square, of course). That way, any finite sum of terms will have a largest term, and the rest of the terms in the sum won't be enough to reach the next square from there.
So, for instance, let $a_1=2$. Then we can let $a_2=5$, because the next square from $5$ is $9$, and $2+5<9$.
From here, $2+5=7$, so the next candidate for $a_3$ is $17$, as the next square up from $17$ is $25$, and you can't reach $25$ from $17$ by adding $2$ and / or $5$.
We now have $2+5+17=24$, so the next candidate is $170$, as the next square from there is $196$, and $2,5,17$ aren't large enough to reach that high.
This can continue indefinitely. The gaps between consecutive squares grow, so you can always find a large enough gap to put the next element of your sequence.
answered Dec 23 '18 at 0:17
Arthur
111k7105186
2
One can, of course, be more compact. For instance, take $a_1=2,a_2=3,a_3=5,a_4=10,ldots$, where each term is the lowest possible number which doesn't allow any sum of terms to be a perfect square. But it quickly becomes tricky to keep track of everything to correctly find the next number in the sequence (and one even has to do an argument similar to the one above to prove that there is always a next term), so I like the approach in the answer better for its simplicity.
– Arthur
Dec 23 '18 at 0:31
add a comment |
Pick any sequence $a_1,a_2,ldots$ which grows fast enough so that $a_1+a_2+ldots a_{n-1}$ is smaller than the distance from $a_n$ up to the next perfect square (and $a_n$ should itself never be a perfect square, of course). That way, any finite sum of terms will have a largest term, and the rest of the terms in the sum won't be enough to reach the next square from there.
So, for instance, let $a_1=2$. Then we can let $a_2=5$, because the next square from $5$ is $9$, and $2+5<9$.
From here, $2+5=7$, so the next candidate for $a_3$ is $17$, as the next square up from $17$ is $25$, and you can't reach $25$ from $17$ by adding $2$ and / or $5$.
We now have $2+5+17=24$, so the next candidate is $170$, as the next square from there is $196$, and $2,5,17$ aren't large enough to reach that high.
This can continue indefinitely. The gaps between consecutive squares grow, so you can always find a large enough gap to put the next element of your sequence.
answered Dec 23 '18 at 0:17
Arthur
111k7105186
2
One can, of course, be more compact. For instance, take $a_1=2,a_2=3,a_3=5,a_4=10,ldots$, where each term is the lowest possible number which doesn't allow any sum of terms to be a perfect square. But it quickly becomes tricky to keep track of everything to correctly find the next number in the sequence (and one even has to do an argument similar to the one above to prove that there is always a next term), so I like the approach in the answer better for its simplicity.
– Arthur
Dec 23 '18 at 0:31
add a comment |
Pick any sequence $a_1,a_2,ldots$ which grows fast enough so that $a_1+a_2+ldots a_{n-1}$ is smaller than the distance from $a_n$ up to the next perfect square (and $a_n$ should itself never be a perfect square, of course). That way, any finite sum of terms will have a largest term, and the rest of the terms in the sum won't be enough to reach the next square from there.
So, for instance, let $a_1=2$. Then we can let $a_2=5$, because the next square from $5$ is $9$, and $2+5<9$.
From here, $2+5=7$, so the next candidate for $a_3$ is $17$, as the next square up from $17$ is $25$, and you can't reach $25$ from $17$ by adding $2$ and / or $5$.
We now have $2+5+17=24$, so the next candidate is $170$, as the next square from there is $196$, and $2,5,17$ aren't large enough to reach that high.
This can continue indefinitely. The gaps between consecutive squares grow, so you can always find a large enough gap to put the next element of your sequence.
answered Dec 23 '18 at 0:17
Arthur
111k7105186
Pick any sequence $a_1,a_2,ldots$ which grows fast enough so that $a_1+a_2+ldots a_{n-1}$ is smaller than the distance from $a_n$ up to the next perfect square (and $a_n$ should itself never be a perfect square, of course). That way, any finite sum of terms will have a largest term, and the rest of the terms in the sum won't be enough to reach the next square from there.
So, for instance, let $a_1=2$. Then we can let $a_2=5$, because the next square from $5$ is $9$, and $2+5<9$.
From here, $2+5=7$, so the next candidate for $a_3$ is $17$, as the next square up from $17$ is $25$, and you can't reach $25$ from $17$ by adding $2$ and / or $5$.
We now have $2+5+17=24$, so the next candidate is $170$, as the next square from there is $196$, and $2,5,17$ aren't large enough to reach that high.
This can continue indefinitely. The gaps between consecutive squares grow, so you can always find a large enough gap to put the next element of your sequence.
answered Dec 23 '18 at 0:17
Arthur
111k7105186
edited Dec 23 '18 at 0:23
answered Dec 23 '18 at 0:17
Arthur
111k7105186
answered Dec 23 '18 at 0:17
Arthur
111k7105186
answered Dec 23 '18 at 0:17
Arthur
111k7105186
111k7105186
2
One can, of course, be more compact. For instance, take $a_1=2,a_2=3,a_3=5,a_4=10,ldots$, where each term is the lowest possible number which doesn't allow any sum of terms to be a perfect square. But it quickly becomes tricky to keep track of everything to correctly find the next number in the sequence (and one even has to do an argument similar to the one above to prove that there is always a next term), so I like the approach in the answer better for its simplicity.
– Arthur
Dec 23 '18 at 0:31
add a comment |
2
One can, of course, be more compact. For instance, take $a_1=2,a_2=3,a_3=5,a_4=10,ldots$, where each term is the lowest possible number which doesn't allow any sum of terms to be a perfect square. But it quickly becomes tricky to keep track of everything to correctly find the next number in the sequence (and one even has to do an argument similar to the one above to prove that there is always a next term), so I like the approach in the answer better for its simplicity.
– Arthur
Dec 23 '18 at 0:31
2
2
One can, of course, be more compact. For instance, take $a_1=2,a_2=3,a_3=5,a_4=10,ldots$, where each term is the lowest possible number which doesn't allow any sum of terms to be a perfect square. But it quickly becomes tricky to keep track of everything to correctly find the next number in the sequence (and one even has to do an argument similar to the one above to prove that there is always a next term), so I like the approach in the answer better for its simplicity.
– Arthur
Dec 23 '18 at 0:31
One can, of course, be more compact. For instance, take $a_1=2,a_2=3,a_3=5,a_4=10,ldots$, where each term is the lowest possible number which doesn't allow any sum of terms to be a perfect square. But it quickly becomes tricky to keep track of everything to correctly find the next number in the sequence (and one even has to do an argument similar to the one above to prove that there is always a next term), so I like the approach in the answer better for its simplicity.
– Arthur
Dec 23 '18 at 0:31
add a comment |
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