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Is $n^2+3n+6$ divisible by 25, where $n$ is a integer?

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1

$begingroup$

If we want $n^2+3n+6$ to be divisible by $25$, it firstly has to be divisible by $5$.
So, let's take a look at couple of cases:
$n=5k: 25k^2+15k+6$. The remainder is $6$, so it's not divisible by $25$.

$n=5k+1: 25k^2+25k+10$. The remainder is $10$, so it is not divisible by $25$.

And so on for $n=5k+2, n=5k+3, n=5k+4.$

1. Is this a good way to do this?

The other way that came to my mind would be following:

Let's say that $n^2+3n+6$ is divisible by $25$: $$n^2+3n+6=25k$$
$$n^2+3n+(6-25k)=0$$
If we solve this equation, we get $$n_1,2=frac-3 pm sqrt5 sqrt20k-32.$$ But $sqrt5$ is a irrational number, so is $n$ irrational number too. This is contradiction, so it's not divisible by $25$.

2. Is this a good way to solve this problem?

number-theory elementary-number-theory divisibility

share|cite|improve this question

edited Mar 11 at 10:21

Gurjinder

554417

asked Mar 11 at 9:53

user389231

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Your second method needs to consider $sqrt20k-3$ too as that might balance the $sqrt5$ and make the overall result rational - in fact it does not, but you have not shown it does not
  $endgroup$
  – Henry
  Mar 11 at 9:58
  

  
  
  
  

add a comment | 

1

$begingroup$

If we want $n^2+3n+6$ to be divisible by $25$, it firstly has to be divisible by $5$.
So, let's take a look at couple of cases:
$n=5k: 25k^2+15k+6$. The remainder is $6$, so it's not divisible by $25$.

$n=5k+1: 25k^2+25k+10$. The remainder is $10$, so it is not divisible by $25$.

And so on for $n=5k+2, n=5k+3, n=5k+4.$

1. Is this a good way to do this?

The other way that came to my mind would be following:

Let's say that $n^2+3n+6$ is divisible by $25$: $$n^2+3n+6=25k$$
$$n^2+3n+(6-25k)=0$$
If we solve this equation, we get $$n_1,2=frac-3 pm sqrt5 sqrt20k-32.$$ But $sqrt5$ is a irrational number, so is $n$ irrational number too. This is contradiction, so it's not divisible by $25$.

2. Is this a good way to solve this problem?

number-theory elementary-number-theory divisibility

share|cite|improve this question

edited Mar 11 at 10:21

Gurjinder

554417

asked Mar 11 at 9:53

user389231

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Your second method needs to consider $sqrt20k-3$ too as that might balance the $sqrt5$ and make the overall result rational - in fact it does not, but you have not shown it does not
  $endgroup$
  – Henry
  Mar 11 at 9:58
  

  
  
  
  

add a comment | 

$begingroup$

If we want $n^2+3n+6$ to be divisible by $25$, it firstly has to be divisible by $5$.
So, let's take a look at couple of cases:
$n=5k: 25k^2+15k+6$. The remainder is $6$, so it's not divisible by $25$.

$n=5k+1: 25k^2+25k+10$. The remainder is $10$, so it is not divisible by $25$.

And so on for $n=5k+2, n=5k+3, n=5k+4.$

1. Is this a good way to do this?

The other way that came to my mind would be following:

Let's say that $n^2+3n+6$ is divisible by $25$: $$n^2+3n+6=25k$$
$$n^2+3n+(6-25k)=0$$
If we solve this equation, we get $$n_1,2=frac-3 pm sqrt5 sqrt20k-32.$$ But $sqrt5$ is a irrational number, so is $n$ irrational number too. This is contradiction, so it's not divisible by $25$.

2. Is this a good way to solve this problem?

number-theory elementary-number-theory divisibility

share|cite|improve this question

edited Mar 11 at 10:21

Gurjinder

554417

asked Mar 11 at 9:53

user389231

$endgroup$

share|cite|improve this question

edited Mar 11 at 10:21

Gurjinder

554417

asked Mar 11 at 9:53

user389231

share|cite|improve this question

edited Mar 11 at 10:21

Gurjinder

554417

asked Mar 11 at 9:53

user389231

edited Mar 11 at 10:21

Gurjinder

554417

edited Mar 11 at 10:21

Gurjinder

554417

2 Answers
2

active

oldest

votes

2

$begingroup$

If $25$ divides $n^2+3n+6,$ the later will be divisible by $5$

$n^2+3n+6equiv n^2-2n+1pmod5$

So, we need $nequiv1pmod5implies n=1+5m$

$(5m+1)^2+3(5m+1)+6=25m^2+25m+10notequiv0pmod25$

share|cite|improve this answer

edited Mar 11 at 13:13

answered Mar 11 at 10:08

lab bhattacharjeelab bhattacharjee

226k15158275

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dietrich, extra dot
  $endgroup$
  – lab bhattacharjee
  Mar 11 at 13:13
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: We have
$$
n^2+3n+6=(n+4)^2 bmod 5
$$

Now look at Bill's answer at this duplicate:

$n^2 + 3n +5$ is not divisible by $121$

share|cite|improve this answer

answered Mar 11 at 10:05

Dietrich BurdeDietrich Burde

80.6k647104

$endgroup$

add a comment | 

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2

$begingroup$

If $25$ divides $n^2+3n+6,$ the later will be divisible by $5$

$n^2+3n+6equiv n^2-2n+1pmod5$

So, we need $nequiv1pmod5implies n=1+5m$

$(5m+1)^2+3(5m+1)+6=25m^2+25m+10notequiv0pmod25$

share|cite|improve this answer

edited Mar 11 at 13:13

answered Mar 11 at 10:08

lab bhattacharjeelab bhattacharjee

226k15158275

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dietrich, extra dot
  $endgroup$
  – lab bhattacharjee
  Mar 11 at 13:13
  

  
  
  
  

add a comment | 

2

$begingroup$

If $25$ divides $n^2+3n+6,$ the later will be divisible by $5$

$n^2+3n+6equiv n^2-2n+1pmod5$

So, we need $nequiv1pmod5implies n=1+5m$

$(5m+1)^2+3(5m+1)+6=25m^2+25m+10notequiv0pmod25$

share|cite|improve this answer

edited Mar 11 at 13:13

answered Mar 11 at 10:08

lab bhattacharjeelab bhattacharjee

226k15158275

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Dietrich, extra dot
  $endgroup$
  – lab bhattacharjee
  Mar 11 at 13:13
  

  
  
  
  

add a comment | 

$begingroup$

If $25$ divides $n^2+3n+6,$ the later will be divisible by $5$

$n^2+3n+6equiv n^2-2n+1pmod5$

So, we need $nequiv1pmod5implies n=1+5m$

$(5m+1)^2+3(5m+1)+6=25m^2+25m+10notequiv0pmod25$

share|cite|improve this answer

edited Mar 11 at 13:13

answered Mar 11 at 10:08

lab bhattacharjeelab bhattacharjee

226k15158275

$endgroup$

share|cite|improve this answer

edited Mar 11 at 13:13

answered Mar 11 at 10:08

lab bhattacharjeelab bhattacharjee

226k15158275

answered Mar 11 at 10:08

lab bhattacharjeelab bhattacharjee

226k15158275

answered Mar 11 at 10:08

lab bhattacharjeelab bhattacharjee

226k15158275

1

$begingroup$

Hint: We have
$$
n^2+3n+6=(n+4)^2 bmod 5
$$

Now look at Bill's answer at this duplicate:

$n^2 + 3n +5$ is not divisible by $121$

share|cite|improve this answer

answered Mar 11 at 10:05

Dietrich BurdeDietrich Burde

80.6k647104

$endgroup$

add a comment | 

1

$begingroup$

Hint: We have
$$
n^2+3n+6=(n+4)^2 bmod 5
$$

Now look at Bill's answer at this duplicate:

$n^2 + 3n +5$ is not divisible by $121$

share|cite|improve this answer

answered Mar 11 at 10:05

Dietrich BurdeDietrich Burde

80.6k647104

$endgroup$

add a comment | 

$begingroup$

Hint: We have
$$
n^2+3n+6=(n+4)^2 bmod 5
$$

Now look at Bill's answer at this duplicate:

$n^2 + 3n +5$ is not divisible by $121$

share|cite|improve this answer

answered Mar 11 at 10:05

Dietrich BurdeDietrich Burde

80.6k647104

$endgroup$

share|cite|improve this answer

answered Mar 11 at 10:05

Dietrich BurdeDietrich Burde

80.6k647104

answered Mar 11 at 10:05

Dietrich BurdeDietrich Burde

80.6k647104

answered Mar 11 at 10:05

Dietrich BurdeDietrich Burde

80.6k647104