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Positive integers have the form $a^2+b^2+c^2+2^d$
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For any positive integer $n$ there seems to be non-negative integers $a,b,c,d$ such that
$$n=a^2+b^2+c^2+2^d.$$
Due to Legendre's three-square theorem a natural number can be represented as the sum of three squares of integers
$$m=a^2+b^2+c^2$$
if and only if $m$ is not of the form $m=4^a(8b+7)$, for integers $a$ and $b$. If the conjecture is true, then for all $ngeq 1$ it must exist a $k$ such that $n-2^k$ is not of the form $m=4^a(8b+7)$.
I want help to prove the conjecture or to find a counter-example.
elementary-number-theory conjectures sums-of-squares
asked 10 hours ago
Lehs
6,77731459
add a comment |Â
up vote
5
down vote
favorite
For any positive integer $n$ there seems to be non-negative integers $a,b,c,d$ such that
$$n=a^2+b^2+c^2+2^d.$$
Due to Legendre's three-square theorem a natural number can be represented as the sum of three squares of integers
$$m=a^2+b^2+c^2$$
if and only if $m$ is not of the form $m=4^a(8b+7)$, for integers $a$ and $b$. If the conjecture is true, then for all $ngeq 1$ it must exist a $k$ such that $n-2^k$ is not of the form $m=4^a(8b+7)$.
I want help to prove the conjecture or to find a counter-example.
elementary-number-theory conjectures sums-of-squares
asked 10 hours ago
Lehs
6,77731459
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
For any positive integer $n$ there seems to be non-negative integers $a,b,c,d$ such that
$$n=a^2+b^2+c^2+2^d.$$
Due to Legendre's three-square theorem a natural number can be represented as the sum of three squares of integers
$$m=a^2+b^2+c^2$$
if and only if $m$ is not of the form $m=4^a(8b+7)$, for integers $a$ and $b$. If the conjecture is true, then for all $ngeq 1$ it must exist a $k$ such that $n-2^k$ is not of the form $m=4^a(8b+7)$.
I want help to prove the conjecture or to find a counter-example.
elementary-number-theory conjectures sums-of-squares
asked 10 hours ago
Lehs
6,77731459
For any positive integer $n$ there seems to be non-negative integers $a,b,c,d$ such that
$$n=a^2+b^2+c^2+2^d.$$
Due to Legendre's three-square theorem a natural number can be represented as the sum of three squares of integers
$$m=a^2+b^2+c^2$$
if and only if $m$ is not of the form $m=4^a(8b+7)$, for integers $a$ and $b$. If the conjecture is true, then for all $ngeq 1$ it must exist a $k$ such that $n-2^k$ is not of the form $m=4^a(8b+7)$.
I want help to prove the conjecture or to find a counter-example.
elementary-number-theory conjectures sums-of-squares
asked 10 hours ago
Lehs
6,77731459
asked 10 hours ago
Lehs
6,77731459
asked 10 hours ago
Lehs
6,77731459
asked 10 hours ago
Lehs
6,77731459
6,77731459
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
We claim that $n$ can be written as $a^2+b^2+c^2+x$ where $xin1,2,4$. Indeed, otherwise, we must have that
$$n-1,n-2,n-4$$
are each of the form $4^a(8b+7)$. However, all such numbers are either $equiv 0$ or $equiv 3bmod 4$, while $n-1,n-2,n-4$ are each members of different residues $bmod 4$. So, it only remains to prove this in the case for which these are not all positive integers, namely $nleq 4$. We have
$$0^2+0^2+0^2+2^0=1$$
$$0^2+0^2+0^2+2^1=2$$
$$0^2+0^2+1^2+2^1=3$$
$$0^2+0^2+0^2+2^2=4.$$
answered 10 hours ago
Carl Schildkraut
8,23711138
or ≡3mod4, while n−1,n−2,n−4 are each members of different residues: How do you know that $a^2+b^2+c^2$ can only be $1$ or $2$ ?
– Diger
9 hours ago
A nice theorem!
– Lehs
9 hours ago
U talking to me?
– Diger
8 hours ago
@Diger $a^2+b^2+c^2$ can be equivalent to $0$ or $3bmod 4$, but using the result that Lehs states in the question, we know the only numbers not representable in that form are those of the form $4^x(8y+7)$ for some nonnegative integers $x,y$. All such numbers are either equivalent to $0$ or $3bmod 4$, so $a^2+b^2+c^2$ reaches all numbers equivalent to $1$ or $2bmod 4$. At least one of $(n-1,n-2,n-4)$ is of this form.
– Carl Schildkraut
8 hours ago
At least one of (n−1,n−2,n−4) is of this form?? I thought all?
– Diger
7 hours ago
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
We claim that $n$ can be written as $a^2+b^2+c^2+x$ where $xin1,2,4$. Indeed, otherwise, we must have that
$$n-1,n-2,n-4$$
are each of the form $4^a(8b+7)$. However, all such numbers are either $equiv 0$ or $equiv 3bmod 4$, while $n-1,n-2,n-4$ are each members of different residues $bmod 4$. So, it only remains to prove this in the case for which these are not all positive integers, namely $nleq 4$. We have
$$0^2+0^2+0^2+2^0=1$$
$$0^2+0^2+0^2+2^1=2$$
$$0^2+0^2+1^2+2^1=3$$
$$0^2+0^2+0^2+2^2=4.$$
answered 10 hours ago
Carl Schildkraut
8,23711138
or ≡3mod4, while n−1,n−2,n−4 are each members of different residues: How do you know that $a^2+b^2+c^2$ can only be $1$ or $2$ ?
– Diger
9 hours ago
A nice theorem!
– Lehs
9 hours ago
U talking to me?
– Diger
8 hours ago
@Diger $a^2+b^2+c^2$ can be equivalent to $0$ or $3bmod 4$, but using the result that Lehs states in the question, we know the only numbers not representable in that form are those of the form $4^x(8y+7)$ for some nonnegative integers $x,y$. All such numbers are either equivalent to $0$ or $3bmod 4$, so $a^2+b^2+c^2$ reaches all numbers equivalent to $1$ or $2bmod 4$. At least one of $(n-1,n-2,n-4)$ is of this form.
– Carl Schildkraut
8 hours ago
At least one of (n−1,n−2,n−4) is of this form?? I thought all?
– Diger
7 hours ago
 |Â
show 1 more comment
up vote
6
down vote
accepted
We claim that $n$ can be written as $a^2+b^2+c^2+x$ where $xin1,2,4$. Indeed, otherwise, we must have that
$$n-1,n-2,n-4$$
are each of the form $4^a(8b+7)$. However, all such numbers are either $equiv 0$ or $equiv 3bmod 4$, while $n-1,n-2,n-4$ are each members of different residues $bmod 4$. So, it only remains to prove this in the case for which these are not all positive integers, namely $nleq 4$. We have
$$0^2+0^2+0^2+2^0=1$$
$$0^2+0^2+0^2+2^1=2$$
$$0^2+0^2+1^2+2^1=3$$
$$0^2+0^2+0^2+2^2=4.$$
answered 10 hours ago
Carl Schildkraut
8,23711138
or ≡3mod4, while n−1,n−2,n−4 are each members of different residues: How do you know that $a^2+b^2+c^2$ can only be $1$ or $2$ ?
– Diger
9 hours ago
A nice theorem!
– Lehs
9 hours ago
U talking to me?
– Diger
8 hours ago
@Diger $a^2+b^2+c^2$ can be equivalent to $0$ or $3bmod 4$, but using the result that Lehs states in the question, we know the only numbers not representable in that form are those of the form $4^x(8y+7)$ for some nonnegative integers $x,y$. All such numbers are either equivalent to $0$ or $3bmod 4$, so $a^2+b^2+c^2$ reaches all numbers equivalent to $1$ or $2bmod 4$. At least one of $(n-1,n-2,n-4)$ is of this form.
– Carl Schildkraut
8 hours ago
At least one of (n−1,n−2,n−4) is of this form?? I thought all?
– Diger
7 hours ago
 |Â
show 1 more comment
up vote
6
down vote
accepted
up vote
6
down vote
accepted
We claim that $n$ can be written as $a^2+b^2+c^2+x$ where $xin1,2,4$. Indeed, otherwise, we must have that
$$n-1,n-2,n-4$$
are each of the form $4^a(8b+7)$. However, all such numbers are either $equiv 0$ or $equiv 3bmod 4$, while $n-1,n-2,n-4$ are each members of different residues $bmod 4$. So, it only remains to prove this in the case for which these are not all positive integers, namely $nleq 4$. We have
$$0^2+0^2+0^2+2^0=1$$
$$0^2+0^2+0^2+2^1=2$$
$$0^2+0^2+1^2+2^1=3$$
$$0^2+0^2+0^2+2^2=4.$$
answered 10 hours ago
Carl Schildkraut
8,23711138
We claim that $n$ can be written as $a^2+b^2+c^2+x$ where $xin1,2,4$. Indeed, otherwise, we must have that
$$n-1,n-2,n-4$$
are each of the form $4^a(8b+7)$. However, all such numbers are either $equiv 0$ or $equiv 3bmod 4$, while $n-1,n-2,n-4$ are each members of different residues $bmod 4$. So, it only remains to prove this in the case for which these are not all positive integers, namely $nleq 4$. We have
$$0^2+0^2+0^2+2^0=1$$
$$0^2+0^2+0^2+2^1=2$$
$$0^2+0^2+1^2+2^1=3$$
$$0^2+0^2+0^2+2^2=4.$$
answered 10 hours ago
Carl Schildkraut
8,23711138
answered 10 hours ago
Carl Schildkraut
8,23711138
answered 10 hours ago
Carl Schildkraut
8,23711138
answered 10 hours ago
Carl Schildkraut
8,23711138
8,23711138
or ≡3mod4, while n−1,n−2,n−4 are each members of different residues: How do you know that $a^2+b^2+c^2$ can only be $1$ or $2$ ?
– Diger
9 hours ago
A nice theorem!
– Lehs
9 hours ago
U talking to me?
– Diger
8 hours ago
@Diger $a^2+b^2+c^2$ can be equivalent to $0$ or $3bmod 4$, but using the result that Lehs states in the question, we know the only numbers not representable in that form are those of the form $4^x(8y+7)$ for some nonnegative integers $x,y$. All such numbers are either equivalent to $0$ or $3bmod 4$, so $a^2+b^2+c^2$ reaches all numbers equivalent to $1$ or $2bmod 4$. At least one of $(n-1,n-2,n-4)$ is of this form.
– Carl Schildkraut
8 hours ago
At least one of (n−1,n−2,n−4) is of this form?? I thought all?
– Diger
7 hours ago
 |Â
show 1 more comment
or ≡3mod4, while n−1,n−2,n−4 are each members of different residues: How do you know that $a^2+b^2+c^2$ can only be $1$ or $2$ ?
– Diger
9 hours ago
A nice theorem!
– Lehs
9 hours ago
U talking to me?
– Diger
8 hours ago
@Diger $a^2+b^2+c^2$ can be equivalent to $0$ or $3bmod 4$, but using the result that Lehs states in the question, we know the only numbers not representable in that form are those of the form $4^x(8y+7)$ for some nonnegative integers $x,y$. All such numbers are either equivalent to $0$ or $3bmod 4$, so $a^2+b^2+c^2$ reaches all numbers equivalent to $1$ or $2bmod 4$. At least one of $(n-1,n-2,n-4)$ is of this form.
– Carl Schildkraut
8 hours ago
At least one of (n−1,n−2,n−4) is of this form?? I thought all?
– Diger
7 hours ago
or ≡3mod4, while n−1,n−2,n−4 are each members of different residues: How do you know that $a^2+b^2+c^2$ can only be $1$ or $2$ ?– Diger
9 hours ago
or ≡3mod4, while n−1,n−2,n−4 are each members of different residues: How do you know that $a^2+b^2+c^2$ can only be $1$ or $2$ ?– Diger
9 hours ago
A nice theorem!
– Lehs
9 hours ago
A nice theorem!
– Lehs
9 hours ago
U talking to me?
– Diger
8 hours ago
U talking to me?
– Diger
8 hours ago
@Diger $a^2+b^2+c^2$ can be equivalent to $0$ or $3bmod 4$, but using the result that Lehs states in the question, we know the only numbers not representable in that form are those of the form $4^x(8y+7)$ for some nonnegative integers $x,y$. All such numbers are either equivalent to $0$ or $3bmod 4$, so $a^2+b^2+c^2$ reaches all numbers equivalent to $1$ or $2bmod 4$. At least one of $(n-1,n-2,n-4)$ is of this form.
– Carl Schildkraut
8 hours ago
@Diger $a^2+b^2+c^2$ can be equivalent to $0$ or $3bmod 4$, but using the result that Lehs states in the question, we know the only numbers not representable in that form are those of the form $4^x(8y+7)$ for some nonnegative integers $x,y$. All such numbers are either equivalent to $0$ or $3bmod 4$, so $a^2+b^2+c^2$ reaches all numbers equivalent to $1$ or $2bmod 4$. At least one of $(n-1,n-2,n-4)$ is of this form.
– Carl Schildkraut
8 hours ago
At least one of (n−1,n−2,n−4) is of this form ?? I thought all?– Diger
7 hours ago
At least one of (n−1,n−2,n−4) is of this form ?? I thought all?– Diger
7 hours ago
 |Â
show 1 more comment
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