Integers which can be written as sum of powers of $2,3$, and $5$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Is it true that every positive integer is the sum of 18 fourth powers of integers?Integers represented by the polynomialLower bound on numbers in “extension” to Lagrange four square theoremIs it known whether any positive integer can be written as the sum of $n$ different squares?Can every element of a finite field be written as a sum of two non-squares?Partition of integers with distinct primesEvery sufficiently large natural number can be written as $p+n_1^2+n_2^2$ for some $n_1,n_2$ - natural numbers and $p$ - prime numberthere is an integer $N_0$ such that for all $Ngeq N_0$ the equation $ax+by=N$ can be solved with both $x$ and $y$ nonnegative integers.Can any integer be written as the sum of $M$ odd powers?Which integers can be written as $x^2+2y^2-3z^2 $?
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Integers which can be written as sum of powers of $2,3$, and $5$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Is it true that every positive integer is the sum of 18 fourth powers of integers?Integers represented by the polynomialLower bound on numbers in “extension” to Lagrange four square theoremIs it known whether any positive integer can be written as the sum of $n$ different squares?Can every element of a finite field be written as a sum of two non-squares?Partition of integers with distinct primesEvery sufficiently large natural number can be written as $p+n_1^2+n_2^2$ for some $n_1,n_2$ - natural numbers and $p$ - prime numberthere is an integer $N_0$ such that for all $Ngeq N_0$ the equation $ax+by=N$ can be solved with both $x$ and $y$ nonnegative integers.Can any integer be written as the sum of $M$ odd powers?Which integers can be written as $x^2+2y^2-3z^2 $?
$begingroup$
Is it true that every sufficiently large integer can be written in the form
$$
2^a3^b5^c+2^d3^e5^f
$$
for some integer $a,b,c,d,e,f ge 0$?
number-theory diophantine-equations
asked Apr 2 at 15:52
Paolo LeonettiPaolo Leonetti
11.3k21550
$endgroup$
add a comment |
$begingroup$
Is it true that every sufficiently large integer can be written in the form
$$
2^a3^b5^c+2^d3^e5^f
$$
for some integer $a,b,c,d,e,f ge 0$?
number-theory diophantine-equations
asked Apr 2 at 15:52
Paolo LeonettiPaolo Leonetti
11.3k21550
$endgroup$
$begingroup$
Given that each exponent is allowed to be $0$, it is true that every sufficiently small integer ($ge 2$) can be so represented.
$endgroup$
– Keith Backman
Apr 2 at 18:19
$begingroup$
@KeithBackman Are you trolling?
$endgroup$
– Paolo Leonetti
Apr 2 at 18:21
$begingroup$
Not at all. I saw Robert Israel's answer, and I tried to find any other small integers that could not be so represented, and found I could construct every integer up to $70$ but not $71$. Just a paper and pencil check on the given answer.
$endgroup$
– Keith Backman
Apr 2 at 18:25
add a comment |
$begingroup$
Is it true that every sufficiently large integer can be written in the form
$$
2^a3^b5^c+2^d3^e5^f
$$
for some integer $a,b,c,d,e,f ge 0$?
number-theory diophantine-equations
asked Apr 2 at 15:52
Paolo LeonettiPaolo Leonetti
11.3k21550
$endgroup$
Is it true that every sufficiently large integer can be written in the form
$$
2^a3^b5^c+2^d3^e5^f
$$
for some integer $a,b,c,d,e,f ge 0$?
number-theory diophantine-equations
number-theory diophantine-equations
asked Apr 2 at 15:52
Paolo LeonettiPaolo Leonetti
11.3k21550
asked Apr 2 at 15:52
Paolo LeonettiPaolo Leonetti
11.3k21550
asked Apr 2 at 15:52
Paolo LeonettiPaolo Leonetti
11.3k21550
asked Apr 2 at 15:52
Paolo LeonettiPaolo Leonetti
11.3k21550
asked Apr 2 at 15:52
Paolo LeonettiPaolo Leonetti
11.3k21550
11.3k21550
$begingroup$
Given that each exponent is allowed to be $0$, it is true that every sufficiently small integer ($ge 2$) can be so represented.
$endgroup$
– Keith Backman
Apr 2 at 18:19
$begingroup$
@KeithBackman Are you trolling?
$endgroup$
– Paolo Leonetti
Apr 2 at 18:21
$begingroup$
Not at all. I saw Robert Israel's answer, and I tried to find any other small integers that could not be so represented, and found I could construct every integer up to $70$ but not $71$. Just a paper and pencil check on the given answer.
$endgroup$
– Keith Backman
Apr 2 at 18:25
add a comment |
$begingroup$
Given that each exponent is allowed to be $0$, it is true that every sufficiently small integer ($ge 2$) can be so represented.
$endgroup$
– Keith Backman
Apr 2 at 18:19
$begingroup$
@KeithBackman Are you trolling?
$endgroup$
– Paolo Leonetti
Apr 2 at 18:21
$begingroup$
Not at all. I saw Robert Israel's answer, and I tried to find any other small integers that could not be so represented, and found I could construct every integer up to $70$ but not $71$. Just a paper and pencil check on the given answer.
$endgroup$
– Keith Backman
Apr 2 at 18:25
$begingroup$
Given that each exponent is allowed to be $0$, it is true that every sufficiently small integer ($ge 2$) can be so represented.
$endgroup$
– Keith Backman
Apr 2 at 18:19
$begingroup$
Given that each exponent is allowed to be $0$, it is true that every sufficiently small integer ($ge 2$) can be so represented.
$endgroup$
– Keith Backman
Apr 2 at 18:19
$begingroup$
@KeithBackman Are you trolling?
$endgroup$
– Paolo Leonetti
Apr 2 at 18:21
$begingroup$
@KeithBackman Are you trolling?
$endgroup$
– Paolo Leonetti
Apr 2 at 18:21
$begingroup$
Not at all. I saw Robert Israel's answer, and I tried to find any other small integers that could not be so represented, and found I could construct every integer up to $70$ but not $71$. Just a paper and pencil check on the given answer.
$endgroup$
– Keith Backman
Apr 2 at 18:25
$begingroup$
Not at all. I saw Robert Israel's answer, and I tried to find any other small integers that could not be so represented, and found I could construct every integer up to $70$ but not $71$. Just a paper and pencil check on the given answer.
$endgroup$
– Keith Backman
Apr 2 at 18:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No, this is not the case. The number of such possible sums $le N$ for large $N$ is far less than $N$.
The number of powers of $2$ (or of $3$ or $5$) up to $N$ is $O(log N)$. Hence the number of products $2^a 3^b 5^c$ up to $N$ is $O((log N)^3)$. The number of sums of pairs of such products is $O((log N)^6)$, which is $o(N)$.
answered Apr 2 at 16:06
FredHFredH
3,6851024
$endgroup$
add a comment |
$begingroup$
No numbers congruent to $71$ or $119$ mod $120$ can be represented in this way.
answered Apr 2 at 16:21
Robert IsraelRobert Israel
332k23222482
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
No, this is not the case. The number of such possible sums $le N$ for large $N$ is far less than $N$.
The number of powers of $2$ (or of $3$ or $5$) up to $N$ is $O(log N)$. Hence the number of products $2^a 3^b 5^c$ up to $N$ is $O((log N)^3)$. The number of sums of pairs of such products is $O((log N)^6)$, which is $o(N)$.
answered Apr 2 at 16:06
FredHFredH
3,6851024
$endgroup$
add a comment |
$begingroup$
No, this is not the case. The number of such possible sums $le N$ for large $N$ is far less than $N$.
The number of powers of $2$ (or of $3$ or $5$) up to $N$ is $O(log N)$. Hence the number of products $2^a 3^b 5^c$ up to $N$ is $O((log N)^3)$. The number of sums of pairs of such products is $O((log N)^6)$, which is $o(N)$.
answered Apr 2 at 16:06
FredHFredH
3,6851024
$endgroup$
add a comment |
$begingroup$
No, this is not the case. The number of such possible sums $le N$ for large $N$ is far less than $N$.
The number of powers of $2$ (or of $3$ or $5$) up to $N$ is $O(log N)$. Hence the number of products $2^a 3^b 5^c$ up to $N$ is $O((log N)^3)$. The number of sums of pairs of such products is $O((log N)^6)$, which is $o(N)$.
answered Apr 2 at 16:06
FredHFredH
3,6851024
$endgroup$
No, this is not the case. The number of such possible sums $le N$ for large $N$ is far less than $N$.
The number of powers of $2$ (or of $3$ or $5$) up to $N$ is $O(log N)$. Hence the number of products $2^a 3^b 5^c$ up to $N$ is $O((log N)^3)$. The number of sums of pairs of such products is $O((log N)^6)$, which is $o(N)$.
answered Apr 2 at 16:06
FredHFredH
3,6851024
answered Apr 2 at 16:06
FredHFredH
3,6851024
answered Apr 2 at 16:06
FredHFredH
3,6851024
answered Apr 2 at 16:06
FredHFredH
3,6851024
3,6851024
add a comment |
add a comment |
$begingroup$
No numbers congruent to $71$ or $119$ mod $120$ can be represented in this way.
answered Apr 2 at 16:21
Robert IsraelRobert Israel
332k23222482
$endgroup$
add a comment |
$begingroup$
No numbers congruent to $71$ or $119$ mod $120$ can be represented in this way.
answered Apr 2 at 16:21
Robert IsraelRobert Israel
332k23222482
$endgroup$
add a comment |
$begingroup$
No numbers congruent to $71$ or $119$ mod $120$ can be represented in this way.
answered Apr 2 at 16:21
Robert IsraelRobert Israel
332k23222482
$endgroup$
No numbers congruent to $71$ or $119$ mod $120$ can be represented in this way.
answered Apr 2 at 16:21
Robert IsraelRobert Israel
332k23222482
answered Apr 2 at 16:21
Robert IsraelRobert Israel
332k23222482
answered Apr 2 at 16:21
Robert IsraelRobert Israel
332k23222482
answered Apr 2 at 16:21
Robert IsraelRobert Israel
332k23222482
332k23222482
add a comment |
add a comment |
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$begingroup$
Given that each exponent is allowed to be $0$, it is true that every sufficiently small integer ($ge 2$) can be so represented.
$endgroup$
– Keith Backman
Apr 2 at 18:19
$begingroup$
@KeithBackman Are you trolling?
$endgroup$
– Paolo Leonetti
Apr 2 at 18:21
$begingroup$
Not at all. I saw Robert Israel's answer, and I tried to find any other small integers that could not be so represented, and found I could construct every integer up to $70$ but not $71$. Just a paper and pencil check on the given answer.
$endgroup$
– Keith Backman
Apr 2 at 18:25