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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?

1

3

$begingroup$

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

share|cite|improve this question

edited Mar 19 at 8:37

user21820

40k544161

asked Mar 19 at 3:54

SaniaSania

406

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, this problem sounds painful.
  $endgroup$
  – Don Thousand
  Mar 19 at 4:08
  

  
  
  
  

add a comment | 

1

3

$begingroup$

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

share|cite|improve this question

edited Mar 19 at 8:37

user21820

40k544161

asked Mar 19 at 3:54

SaniaSania

406

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wow, this problem sounds painful.
  $endgroup$
  – Don Thousand
  Mar 19 at 4:08
  

  
  
  
  

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

share|cite|improve this question

edited Mar 19 at 8:37

user21820

40k544161

asked Mar 19 at 3:54

SaniaSania

406

$endgroup$

share|cite|improve this question

edited Mar 19 at 8:37

user21820

40k544161

asked Mar 19 at 3:54

SaniaSania

406

share|cite|improve this question

edited Mar 19 at 8:37

user21820

40k544161

asked Mar 19 at 3:54

SaniaSania

406

edited Mar 19 at 8:37

user21820

40k544161

edited Mar 19 at 8:37

user21820

40k544161

asked Mar 19 at 3:54

SaniaSania

406

asked Mar 19 at 3:54

SaniaSania

406

2 Answers
2

active

oldest

votes

3

$begingroup$

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 $N+1 equiv 0 pmod {2^2}, N+2 equiv 0 pmod {3^3}, N+3 equiv 0 pmod {5^4}ldots$

share|cite|improve this answer

answered Mar 19 at 4:13

Ross MillikanRoss Millikan

301k24200375

$endgroup$

add a comment | 

3

$begingroup$

By the Chinese remainder theorem, there is an integer $n$ such that $nequiv -k (text{mod} p_k^{k+1})$ for all $k=1,2,dots 134$, (where $p_k$ is the $k^text{th}$ smallest prime). Then $n, n+1, dots, n+134$ satisfy the required condition.

share|cite|improve this answer

answered Mar 19 at 4:18

Poon LeviPoon Levi

51638

$endgroup$

add a comment | 

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3

$begingroup$

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 $N+1 equiv 0 pmod {2^2}, N+2 equiv 0 pmod {3^3}, N+3 equiv 0 pmod {5^4}ldots$

share|cite|improve this answer

answered Mar 19 at 4:13

Ross MillikanRoss Millikan

301k24200375

$endgroup$

add a comment | 

3

$begingroup$

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 $N+1 equiv 0 pmod {2^2}, N+2 equiv 0 pmod {3^3}, N+3 equiv 0 pmod {5^4}ldots$

share|cite|improve this answer

answered Mar 19 at 4:13

Ross MillikanRoss Millikan

301k24200375

$endgroup$

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 $N+1 equiv 0 pmod {2^2}, N+2 equiv 0 pmod {3^3}, N+3 equiv 0 pmod {5^4}ldots$

share|cite|improve this answer

answered Mar 19 at 4:13

Ross MillikanRoss Millikan

301k24200375

$endgroup$

share|cite|improve this answer

answered Mar 19 at 4:13

Ross MillikanRoss Millikan

301k24200375

answered Mar 19 at 4:13

Ross MillikanRoss Millikan

301k24200375

answered Mar 19 at 4:13

Ross MillikanRoss Millikan

301k24200375

3

$begingroup$

By the Chinese remainder theorem, there is an integer $n$ such that $nequiv -k (text{mod} p_k^{k+1})$ for all $k=1,2,dots 134$, (where $p_k$ is the $k^text{th}$ smallest prime). Then $n, n+1, dots, n+134$ satisfy the required condition.

share|cite|improve this answer

answered Mar 19 at 4:18

Poon LeviPoon Levi

51638

$endgroup$

add a comment | 

3

$begingroup$

By the Chinese remainder theorem, there is an integer $n$ such that $nequiv -k (text{mod} p_k^{k+1})$ for all $k=1,2,dots 134$, (where $p_k$ is the $k^text{th}$ smallest prime). Then $n, n+1, dots, n+134$ satisfy the required condition.

share|cite|improve this answer

answered Mar 19 at 4:18

Poon LeviPoon Levi

51638

$endgroup$

add a comment | 

$begingroup$

By the Chinese remainder theorem, there is an integer $n$ such that $nequiv -k (text{mod} p_k^{k+1})$ for all $k=1,2,dots 134$, (where $p_k$ is the $k^text{th}$ smallest prime). Then $n, n+1, dots, n+134$ satisfy the required condition.

share|cite|improve this answer

answered Mar 19 at 4:18

Poon LeviPoon Levi

51638

$endgroup$

share|cite|improve this answer

answered Mar 19 at 4:18

Poon LeviPoon Levi

51638

answered Mar 19 at 4:18

Poon LeviPoon Levi

51638

answered Mar 19 at 4:18

Poon LeviPoon Levi

51638