Prove that there exist $135$ consecutive positive integers so that the $n$th least is divisible by a perfect $n$th power greater than $1$product of six consecutive integers being a perfect squareCan both $x^2 + y+2$ and $y^2+4x$ be squares?$n mid k^2 land n+1 mid l^3 land n+2 mid m^4 to n=?$Set whose average of subsets is always square (cube, etc.)Prove that there exist two integers such that i - j is divisible by n.Number Theory Homework: Find 3 consecutive integers…Question from Mathcounts competitionProve that there are no positive integers $x$ and $y$ such that $x^3 + y^3 = 10^3$.odd integers that are divisible by a perfect cubeCan I complete the euclidean case without elliptic-curve-theory?
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Prove that there exist $135$ consecutive positive integers so that the $n$th least is divisible by a perfect $n$th power greater than $1$
product of six consecutive integers being a perfect squareCan both $x^2 + y+2$ and $y^2+4x$ be squares?$n mid k^2 land n+1 mid l^3 land n+2 mid m^4 to n=?$Set whose average of subsets is always square (cube, etc.)Prove that there exist two integers such that i - j is divisible by n.Number Theory Homework: Find 3 consecutive integers…Question from Mathcounts competitionProve that there are no positive integers $x$ and $y$ such that $x^3 + y^3 = 10^3$.odd integers that are divisible by a perfect cubeCan I complete the euclidean case without elliptic-curve-theory?
$begingroup$
Prove that there exist 135 consecutive positive integers so that the second least is divisible by a perfect square $> 1$, the third least is divisible by a perfect cube $> 1$, the fourth least is divisible by a perfect fourth power $> 1$, and so on.
How should I go about doing this?
I thought perhaps I should use Fermat's little theorem, or its corollary?
Thanks!
elementary-number-theory divisibility
edited 16 hours ago
user21820
39.6k543156
asked 21 hours ago
SaniaSania
314
New contributor
Sania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
Prove that there exist 135 consecutive positive integers so that the second least is divisible by a perfect square $> 1$, the third least is divisible by a perfect cube $> 1$, the fourth least is divisible by a perfect fourth power $> 1$, and so on.
How should I go about doing this?
I thought perhaps I should use Fermat's little theorem, or its corollary?
Thanks!
elementary-number-theory divisibility
edited 16 hours ago
user21820
39.6k543156
asked 21 hours ago
SaniaSania
314
New contributor
Sania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Wow, this problem sounds painful.
$endgroup$
– Don Thousand
21 hours ago
add a comment |
$begingroup$
Prove that there exist 135 consecutive positive integers so that the second least is divisible by a perfect square $> 1$, the third least is divisible by a perfect cube $> 1$, the fourth least is divisible by a perfect fourth power $> 1$, and so on.
How should I go about doing this?
I thought perhaps I should use Fermat's little theorem, or its corollary?
Thanks!
elementary-number-theory divisibility
edited 16 hours ago
user21820
39.6k543156
asked 21 hours ago
SaniaSania
314
New contributor
Sania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Prove that there exist 135 consecutive positive integers so that the second least is divisible by a perfect square $> 1$, the third least is divisible by a perfect cube $> 1$, the fourth least is divisible by a perfect fourth power $> 1$, and so on.
How should I go about doing this?
I thought perhaps I should use Fermat's little theorem, or its corollary?
Thanks!
elementary-number-theory divisibility
elementary-number-theory divisibility
edited 16 hours ago
user21820
39.6k543156
asked 21 hours ago
SaniaSania
314
New contributor
Sania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 16 hours ago
user21820
39.6k543156
asked 21 hours ago
SaniaSania
314
New contributor
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Check out our Code of Conduct.
edited 16 hours ago
user21820
39.6k543156
edited 16 hours ago
user21820
39.6k543156
edited 16 hours ago
user21820
39.6k543156
39.6k543156
asked 21 hours ago
SaniaSania
314
New contributor
Sania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 21 hours ago
SaniaSania
314
asked 21 hours ago
SaniaSania
314
314
New contributor
Sania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sania is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Wow, this problem sounds painful.
$endgroup$
– Don Thousand
21 hours ago
add a comment |
$begingroup$
Wow, this problem sounds painful.
$endgroup$
– Don Thousand
21 hours ago
$begingroup$
Wow, this problem sounds painful.
$endgroup$
– Don Thousand
21 hours ago
$begingroup$
Wow, this problem sounds painful.
$endgroup$
– Don Thousand
21 hours ago
add a comment |
2 Answers
2
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oldest
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$begingroup$
Use the Chinese Remainder Theorem. Pick $134$ distinct primes. The perfect square is the square of the first, the cube is the cube of the second, and so on. All your moduli are distinct, so CRT guarantees a solution. If you use the smallest primes in order and $N$ is the least of your $135$ numbers, you have $N+1 equiv 0 pmod 2^2, N+2 equiv 0 pmod 3^3, N+3 equiv 0 pmod 5^4ldots$
answered 21 hours ago
Ross MillikanRoss Millikan
299k24200374
$endgroup$
add a comment |
$begingroup$
By the Chinese remainder theorem, there is an integer $n$ such that $nequiv -k (textmod p_k^k+1)$ for all $k=1,2,dots 134$, (where $p_k$ is the $k^textth$ smallest prime). Then $n, n+1, dots, n+134$ satisfy the required condition.
answered 21 hours ago
Poon LeviPoon Levi
48638
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Use the Chinese Remainder Theorem. Pick $134$ distinct primes. The perfect square is the square of the first, the cube is the cube of the second, and so on. All your moduli are distinct, so CRT guarantees a solution. If you use the smallest primes in order and $N$ is the least of your $135$ numbers, you have $N+1 equiv 0 pmod 2^2, N+2 equiv 0 pmod 3^3, N+3 equiv 0 pmod 5^4ldots$
answered 21 hours ago
Ross MillikanRoss Millikan
299k24200374
$endgroup$
add a comment |
$begingroup$
Use the Chinese Remainder Theorem. Pick $134$ distinct primes. The perfect square is the square of the first, the cube is the cube of the second, and so on. All your moduli are distinct, so CRT guarantees a solution. If you use the smallest primes in order and $N$ is the least of your $135$ numbers, you have $N+1 equiv 0 pmod 2^2, N+2 equiv 0 pmod 3^3, N+3 equiv 0 pmod 5^4ldots$
answered 21 hours ago
Ross MillikanRoss Millikan
299k24200374
$endgroup$
add a comment |
$begingroup$
Use the Chinese Remainder Theorem. Pick $134$ distinct primes. The perfect square is the square of the first, the cube is the cube of the second, and so on. All your moduli are distinct, so CRT guarantees a solution. If you use the smallest primes in order and $N$ is the least of your $135$ numbers, you have $N+1 equiv 0 pmod 2^2, N+2 equiv 0 pmod 3^3, N+3 equiv 0 pmod 5^4ldots$
answered 21 hours ago
Ross MillikanRoss Millikan
299k24200374
$endgroup$
Use the Chinese Remainder Theorem. Pick $134$ distinct primes. The perfect square is the square of the first, the cube is the cube of the second, and so on. All your moduli are distinct, so CRT guarantees a solution. If you use the smallest primes in order and $N$ is the least of your $135$ numbers, you have $N+1 equiv 0 pmod 2^2, N+2 equiv 0 pmod 3^3, N+3 equiv 0 pmod 5^4ldots$
answered 21 hours ago
Ross MillikanRoss Millikan
299k24200374
answered 21 hours ago
Ross MillikanRoss Millikan
299k24200374
answered 21 hours ago
Ross MillikanRoss Millikan
299k24200374
answered 21 hours ago
Ross MillikanRoss Millikan
299k24200374
299k24200374
add a comment |
add a comment |
$begingroup$
By the Chinese remainder theorem, there is an integer $n$ such that $nequiv -k (textmod p_k^k+1)$ for all $k=1,2,dots 134$, (where $p_k$ is the $k^textth$ smallest prime). Then $n, n+1, dots, n+134$ satisfy the required condition.
answered 21 hours ago
Poon LeviPoon Levi
48638
$endgroup$
add a comment |
$begingroup$
By the Chinese remainder theorem, there is an integer $n$ such that $nequiv -k (textmod p_k^k+1)$ for all $k=1,2,dots 134$, (where $p_k$ is the $k^textth$ smallest prime). Then $n, n+1, dots, n+134$ satisfy the required condition.
answered 21 hours ago
Poon LeviPoon Levi
48638
$endgroup$
add a comment |
$begingroup$
By the Chinese remainder theorem, there is an integer $n$ such that $nequiv -k (textmod p_k^k+1)$ for all $k=1,2,dots 134$, (where $p_k$ is the $k^textth$ smallest prime). Then $n, n+1, dots, n+134$ satisfy the required condition.
answered 21 hours ago
Poon LeviPoon Levi
48638
$endgroup$
By the Chinese remainder theorem, there is an integer $n$ such that $nequiv -k (textmod p_k^k+1)$ for all $k=1,2,dots 134$, (where $p_k$ is the $k^textth$ smallest prime). Then $n, n+1, dots, n+134$ satisfy the required condition.
answered 21 hours ago
Poon LeviPoon Levi
48638
answered 21 hours ago
Poon LeviPoon Levi
48638
answered 21 hours ago
Poon LeviPoon Levi
48638
answered 21 hours ago
Poon LeviPoon Levi
48638
48638
add a comment |
add a comment |
Sania is a new contributor. Be nice, and check out our Code of Conduct.
Sania is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Wow, this problem sounds painful.
$endgroup$
– Don Thousand
21 hours ago