Is proof correct?

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January 13th 2006, 03:15 AM
DenMac21

Is proof correct?

I have to prove that $a^5-a$ is dividable with 30 for every integer value of $a$ .

My proof is:
$a^5-a=a(a^2-1)(a^2+1)$ and $30 = 2(2^2-1)(2^2+1)$ .
So if we put $a$ instead of 2 (or 2 instead of a) the proof is proven.

I have found in book much complicated solution but this seems ok to me.

Am I right?

P.S. Can someone tell me is there some guide to writing math text in messages on this forum because I couldn't find it.
January 13th 2006, 03:30 AM
TD!

No, you've proven it for a = 2 but not for all a.
January 13th 2006, 04:08 AM
CaptainBlack

Quote:

Originally Posted by DenMac21

I have to prove that $a^5-a$ is dividable with 30 for every integer value of $a$ .

My proof is:
$a^5-a=a(a^2-1)(a^2+1)$ and $30 = 2(2^2-1)(2^2+1)$ .
So if we put $a$ instead of 2 (or 2 instead of a) the proof is proven.

I have found in book much complicated solution but this seems ok to me.

Am I right?

It would help if you told us what level course this
is related to, otherwise you risk getting a totally
incomprehensible explanation of how to prove the
result.

Quote:

P.S. Can someone tell me is there some guide to writing math text in messages on this forum because I couldn't find it.

What you want is the LaTeX guide, its in the following
thread:

http://www.mathhelpforum.com/math-he...read.php?t=266

RonL
January 13th 2006, 04:18 AM
DenMac21

I am learning Algebra for high-school.
January 13th 2006, 06:34 AM
CaptainBlack

[QUOTE=DenMac21]I have to prove that $a^5-a$ is dividable with 30 for every integer value of $a$ .

/[QUOTE]

I'm not sure if this qualifies as high school maths but here goes:

(read $|$ as divides so $a|b$ means $a$ divides $b$ .

if $30|(a^5-a)$ then $5|(a^5-a)$ and $6|(a^5-a)$ , also if
$5|(a^5-a)$ and $6|(a^5-a)$ then $30|(a^5-a)$ because $5$
and $6$ have no common factor.

So we need only show that for any positive integer $a$
that $5|(a^5-a)$ and $6|(a^5-a)$ .

Now any positive integer $a$ may be written

$a=k.5+r$ ,

for some positive integer $k$ and remainder $r=0, 2,3, \mbox{ or }4$ .

Now:

$a^5-a=5.K+r^5 - 5.k-r$ ,

for some positive integer $K$ . so if:

$5|(a^5-a)$

then:

$5|(r^5-r)$

where $r=0, 2,3, \mbox{ or }4$ , and these five cases can be
checked by hand.

Now repeat the same type of argument for $6$ in place
of $5$ to complete the proof.

RonL
January 13th 2006, 09:03 AM
ThePerfectHacker

Just like CaptainBlack said I do not know what high school math is but try it like this:
Show that if it works for $n$ then it must work for $n+1$ . Meaning if it works for 1 it must work for 2 but if it works for 2 then it works for 3 but if it works for 3 it must work for 4.... (This is called induction).
Thus, let $n^5-n$ be divisible by 30.
Then show that $(n+1)^5-(n+1)$ is divisible by 30. Use the binomial theorem
$(n+1)^5-(n+1)=n^5+5n^4+10n^3+10n^2+5n+1-n-1$
Thus,
$(n^5-n)+5n^4+10n^3+10n^2+5n$
But, $n^5-n$ is divisible by 30,
thus,
$5n^4+10n^3+10n^2+5n$ if divisible by 30 then the whole expression is divisible by 30.
But
$5n^4+10n^3+10n^2+5n$ is divisible by 5
$n^4+2n^3+2n^3+n$ . Factor,
$n(n^3+2n^2+2n+1)$ .

Notice if this is true for $n=1,2,3,4,5,6$ thus, it is true for all numbers. Because 7 is considered as 1 because it leaves the same remainder. 8 is considered as 2 because it leaves the same remainder....
Q.E.D.

In a similar way you could have checked all numbers from 1 to 30 in the problem by the same reason that 31 is considered as 1 in the last paragraph. But I did it this way because it is faster and leaves less computation.
January 13th 2006, 09:21 AM
CaptainBlack

Quote:

Originally Posted by ThePerfectHacker

Just like CaptainBlack said I do not know what high school math is but try it like this:
Show that if it works for $n$ then it must work for $n+1$ . Meaning if it works for 1 it must work for 2 but if it works for 2 then it works for 3 but if it works for 3 it must work for 4.... (This is called induction).

I was deliberately avoiding induction, as I suspect its
not high school maths :eek:

Fermat's little theorem is ruled out for the same reason :cool:

RonL
January 13th 2006, 10:15 AM
DenMac21

This is the proof from the book which is (at the moment) most understable to me:

$a^5-a=(a-1)(a)(a+1)(a^2+1)=$
$(a-1)a(a+1)(a^2-4)+5)=$
$(a-2)(a-1)a(a+1)(a+2) + 5(a-1)a(a+1)$

$(a-2)(a-1)a(a+1)(a+2)$ are five successive integer numbers and they are divisible with 30.
$5(a-1)a(a+1)$ have as factors 5 and three successive integer numbers which product is 6 so also this expression is divisible with 30.

So, both expressions are divisible with 30 and their sum is then also divisible with 30.
January 13th 2006, 10:16 AM
CaptainBlack

Quote:

Originally Posted by ThePerfectHacker

Just like CaptainBlack said I do not know what high school math is but try it like this:
Show that if it works for $n$ then it must work for $n+1$ . Meaning if it works for 1 it must work for 2 but if it works for 2 then it works for 3 but if it works for 3 it must work for 4.... (This is called induction).
Thus, let $n^5-n$ be divisible by 30.
Then show that $(n+1)^5-(n+1)$ is divisible by 30. Use the binomial theorem
$(n+1)^5-(n+1)=n^5+5n^4+10n^3+10n^2+5n+1-n-1$
Thus,
$(n^5-n)+5n^4+10n^3+10n^2+5n$
But, $n^5-n$ is divisible by 30,
thus,
$5n^4+10n^3+10n^2+5n$ if divisible by 30 then the whole expression is divisible by 30.
But
$5n^4+10n^3+10n^2+5n$ is divisible by 5
$n^4+2n^3+2n^3+n$ . Factor,
$n(n^3+2n^2+2n+1)$ .

You have lost me around here, presumably you are trying to show that

$n(n^3+2n^2+2n+1)$ .

is divisible by 6 (or 2 and 3), but I can't follow how you are doing so :(

Quote:

Notice if this is true for $n=1,2,3,4,5,6$ thus, it is true for all numbers. Because 7 is considered as 1 because it leaves the same remainder. 8 is considered as 2 because it leaves the same remainder....
Q.E.D.

In a similar way you could have checked all numbers from 1 to 30 in the problem by the same reason that 31 is considered as 1 in the last paragraph. But I did it this way because it is faster and leaves less computation.

RonL
January 13th 2006, 10:21 AM
TD!

Quote:

Originally Posted by CaptainBlack

You have lost me around here, presumably you are trying to show that

$n(n^3+2n^2+2n+1)$ .

is divisible by 6 (or 2 and 3), but I can't follow how you are doing so :(

RonL

Just manually filling in n = 1 -> 6.
January 13th 2006, 10:22 AM
TD!

Quote:

Originally Posted by DenMac21

This is the proof from the book which is (at the moment) most understable to me:

$a^5-a=(a-1)(a)(a+1)(a^2+1)=$
$(a-1)a(a+1)(a^2-4)+5)=$
$(a-2)(a-1)a(a+1)(a+2) + 5(a-1)a(a+1)$

$(a-2)(a-1)a(a+1)(a+2)$ are five successive integer numbers and they are divisible with 30.
$5(a-1)a(a+1)$ have as factors 5 and three successive integer numbers which product is 6 so also this expression is divisible with 30.

So, both expressions are divisible with 30 and their sum is then also divisible with 30.

The proof is fine but you have to come up with the little 'trick' in the second step of course :)
January 13th 2006, 10:26 AM
CaptainBlack

Quote:

Originally Posted by TD!

Just manually filling in n = 1 -> 6.

Well that would do it as it's true, but the words don't seem to make that
clear.

RonL
January 13th 2006, 10:30 AM
TD!

Quote:

Originally Posted by CaptainBlack

Well that would do it as it's true, but the words don't seem to make that
clear.

Perhaps "check" would've been better than "notice", but he meant to say that here. It was also implied a bit in his last paragraph.

Quote:

Originally Posted by ThePerfectHacker

Notice if this is true for $n=1,2,3,4,5,6$ thus, it is true for all numbers. Because 7 is considered as 1 because it leaves the same remainder. 8 is considered as 2 because it leaves the same remainder....
Q.E.D.

In a similar way you could have checked all numbers from 1 to 30 in the problem by the same reason that 31 is considered as 1 in the last paragraph. But I did it this way because it is faster and leaves less computation.
January 13th 2006, 10:35 AM
CaptainBlack

Quote:

Originally Posted by DenMac21

This is the proof from the book which is (at the moment) most understable to me:

$a^5-a=(a-1)(a)(a+1)(a^2+1)=$
$(a-1)a(a+1)(a^2-4)+5)=$
$(a-2)(a-1)a(a+1)(a+2) + 5(a-1)a(a+1)$

$(a-2)(a-1)a(a+1)(a+2)$ are five successive integer numbers and they are divisible with 30.

Seems to me that you might have to prove this last assertion (not too
difficult) unless its a result you are expected to know. Same comment
applies to similar assertion below.

Quote:

$5(a-1)a(a+1)$ have as factors 5 and three successive integer numbers which product is 6 so also this expression is divisible with 30.

So, both expressions are divisible with 30 and their sum is then also divisible with 30.

RonL