Does bounded version of Lagrange's 4-square theorem follow from this result?
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Let $S_4(n)$ be the number of ways of representing $n$ as a sum of 4 integer squares and suppose I can give you a proof of the result that $sum_{i=1}^nS_4(i) sim frac{pi^2}{2}n^2$.
Is this result enough to deduce that beyond some $N$, every integer can be represented as the sum of 4 squares (given that we are "weakly" suggesting that $sum_{i=1}^nS_4(i) >> n$)?
number-theory proof-verification asymptotics sums-of-squares
asked Dec 15 '18 at 16:13
Isky MathewsIsky Mathews
903314
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add a comment |
$begingroup$
Let $S_4(n)$ be the number of ways of representing $n$ as a sum of 4 integer squares and suppose I can give you a proof of the result that $sum_{i=1}^nS_4(i) sim frac{pi^2}{2}n^2$.
Is this result enough to deduce that beyond some $N$, every integer can be represented as the sum of 4 squares (given that we are "weakly" suggesting that $sum_{i=1}^nS_4(i) >> n$)?
number-theory proof-verification asymptotics sums-of-squares
asked Dec 15 '18 at 16:13
Isky MathewsIsky Mathews
903314
$endgroup$
1
$begingroup$
It's not true that $S_4(n)simfrac{pi^2}{2}n^2$. What is true is that $sum_{k=1}^nS_4(k)simfrac{pi^2}{2}n^2$.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 16:16
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Ah yes, I meant that! Let me just update this.
$endgroup$
– Isky Mathews
Dec 15 '18 at 17:12
$begingroup$
@LordSharktheUnknown: Can you answer the question, though? I know from experience on this site that you're quite a capable individual...
$endgroup$
– Isky Mathews
Dec 15 '18 at 17:13
$begingroup$
You need more than that asymptotic.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 19:13
add a comment |
$begingroup$
Let $S_4(n)$ be the number of ways of representing $n$ as a sum of 4 integer squares and suppose I can give you a proof of the result that $sum_{i=1}^nS_4(i) sim frac{pi^2}{2}n^2$.
Is this result enough to deduce that beyond some $N$, every integer can be represented as the sum of 4 squares (given that we are "weakly" suggesting that $sum_{i=1}^nS_4(i) >> n$)?
number-theory proof-verification asymptotics sums-of-squares
asked Dec 15 '18 at 16:13
Isky MathewsIsky Mathews
903314
$endgroup$
Let $S_4(n)$ be the number of ways of representing $n$ as a sum of 4 integer squares and suppose I can give you a proof of the result that $sum_{i=1}^nS_4(i) sim frac{pi^2}{2}n^2$.
Is this result enough to deduce that beyond some $N$, every integer can be represented as the sum of 4 squares (given that we are "weakly" suggesting that $sum_{i=1}^nS_4(i) >> n$)?
number-theory proof-verification asymptotics sums-of-squares
number-theory proof-verification asymptotics sums-of-squares
asked Dec 15 '18 at 16:13
Isky MathewsIsky Mathews
903314
asked Dec 15 '18 at 16:13
Isky MathewsIsky Mathews
903314
edited Dec 15 '18 at 21:39
Isky Mathews
asked Dec 15 '18 at 16:13
Isky MathewsIsky Mathews
903314
asked Dec 15 '18 at 16:13
Isky MathewsIsky Mathews
903314
asked Dec 15 '18 at 16:13
Isky MathewsIsky Mathews
903314
903314
1
$begingroup$
It's not true that $S_4(n)simfrac{pi^2}{2}n^2$. What is true is that $sum_{k=1}^nS_4(k)simfrac{pi^2}{2}n^2$.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 16:16
$begingroup$
Ah yes, I meant that! Let me just update this.
$endgroup$
– Isky Mathews
Dec 15 '18 at 17:12
$begingroup$
@LordSharktheUnknown: Can you answer the question, though? I know from experience on this site that you're quite a capable individual...
$endgroup$
– Isky Mathews
Dec 15 '18 at 17:13
$begingroup$
You need more than that asymptotic.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 19:13
add a comment |
1
$begingroup$
It's not true that $S_4(n)simfrac{pi^2}{2}n^2$. What is true is that $sum_{k=1}^nS_4(k)simfrac{pi^2}{2}n^2$.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 16:16
$begingroup$
Ah yes, I meant that! Let me just update this.
$endgroup$
– Isky Mathews
Dec 15 '18 at 17:12
$begingroup$
@LordSharktheUnknown: Can you answer the question, though? I know from experience on this site that you're quite a capable individual...
$endgroup$
– Isky Mathews
Dec 15 '18 at 17:13
$begingroup$
You need more than that asymptotic.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 19:13
1
1
$begingroup$
It's not true that $S_4(n)simfrac{pi^2}{2}n^2$. What is true is that $sum_{k=1}^nS_4(k)simfrac{pi^2}{2}n^2$.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 16:16
$begingroup$
It's not true that $S_4(n)simfrac{pi^2}{2}n^2$. What is true is that $sum_{k=1}^nS_4(k)simfrac{pi^2}{2}n^2$.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 16:16
$begingroup$
Ah yes, I meant that! Let me just update this.
$endgroup$
– Isky Mathews
Dec 15 '18 at 17:12
$begingroup$
Ah yes, I meant that! Let me just update this.
$endgroup$
– Isky Mathews
Dec 15 '18 at 17:12
$begingroup$
@LordSharktheUnknown: Can you answer the question, though? I know from experience on this site that you're quite a capable individual...
$endgroup$
– Isky Mathews
Dec 15 '18 at 17:13
$begingroup$
@LordSharktheUnknown: Can you answer the question, though? I know from experience on this site that you're quite a capable individual...
$endgroup$
– Isky Mathews
Dec 15 '18 at 17:13
$begingroup$
You need more than that asymptotic.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 19:13
$begingroup$
You need more than that asymptotic.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 19:13
add a comment |
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1
$begingroup$
It's not true that $S_4(n)simfrac{pi^2}{2}n^2$. What is true is that $sum_{k=1}^nS_4(k)simfrac{pi^2}{2}n^2$.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 16:16
$begingroup$
Ah yes, I meant that! Let me just update this.
$endgroup$
– Isky Mathews
Dec 15 '18 at 17:12
$begingroup$
@LordSharktheUnknown: Can you answer the question, though? I know from experience on this site that you're quite a capable individual...
$endgroup$
– Isky Mathews
Dec 15 '18 at 17:13
$begingroup$
You need more than that asymptotic.
$endgroup$
– Lord Shark the Unknown
Dec 15 '18 at 19:13