Help me learn my fellow Brilliantians !!!

I am a newbie at Olympiad level mathematics, so I bought some books to help me, but most of the time I find myself stuck at some questions, which frustrates me very much and then leaves me hopeless, I was hoping if you guys might help me. So I'll provide those questions in this discussion, I hope you'll all help me.

If $m,n,p,q$ are non negative integers prove that, $\displaystyle \sum_{m=0}^{q} (n-m)\dfrac {(p+m)!}{m!}= \dfrac {(p+q+1)!}{q!} \cdot (\dfrac {n}{p+1}-\dfrac{q}{p+2})$
Prove $\dfrac{k^7}{7}+\dfrac{k^5}{5}+\dfrac{2k^3}{3}-\dfrac{k}{105}$ is an positive integer for all positive integers $k$ .
If $m>n$ prove that $a^{2^ n}+1$ is a divisor of $a^{2^m}-1$ . Find LCM of $(a^{2^n}+1, a^{2^m}-1)$ .

Thanks in advance.

#HelpMe! #Proofs #MathProblem #Math

Easy Math Editor

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`2 \times 3`	$2 \times 3$
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Comments

I don't really get what your second question means. Do you mean that the expression given will equal to $k$ or other meaning?

敬全钟 - 7 years, 6 months ago

I am really sorry, I made a typo, I have corrected it now.

Aejeth Lord - 7 years, 6 months ago

If it is interpreted as "find all k such that this polynomial is equal to k when evaluated with k" then we can easily arrive at $k = -1, 0, 1$ , and these are unique solutions. But I find that the question is a bit more complicated than this.

Michael Tong - 7 years, 6 months ago

Actually, I believe he is asking to prove that the polynomial is equal to a positive integer for all positive integral $k$ , In which case the proof is as follows:

The expression is equivalent to $\frac{1}{105}(15k^7+21k^5+70k^3-k)$ , so if we can show that $15k^7+21k^5+70k^3-k$ is divisible by $105=3*5*7$ for all positive integral $k$ then we are done. Consider the expression modulo $3$ , $5$ , and $7$ . We have the equivalences $k^3-k \equiv 0 (mod 3)$ , $k^5-k \equiv 0 (mod 5)$ , and $k^7-k \equiv 0 (mod 7)$ , which are all true for all positive integers $k$ by Fermat's little theorem, so the expression is divisible by $105$ .

Logan Dymond - 7 years, 6 months ago

Q.3.i have seen this same problem some times ago but with the only difference, there is it mentioned that $a,m,n$ are positive integer so i'm writing my solution assuming $a,m,n$ are positive integer.

we can write,

$(a^{2^{m}}-1)=(a^{2^{m-1}}+1)(a^{2^{m-2}}+1)(a^{2^{m-3}}+1)\dots(a^2+1)(a+1)(a-1)$

let, $n=(m-q)$ where $q\in{{1,2,3,\dots,(m-1)}}$

since, $m>n$ and $n$ is positive

so it is clear that for any value of q, $(a^{2{^n}}+1)$ is a factor of $(a^{2^{m}}-1)$

hence, $(a^{2^{n}}+1)$ is a divisor of $(a^{2^{m}}-1)$ if $m>n$

since, $(a^{2^{n}}+1)$ is a divisor of $(a^{2^{m}}-1)$ so LCM of this two = $(a^{2^{m}}-1)$

Gypsy Singer - 7 years, 6 months ago

the second question can be done using induction and binomial theroem

Siddharth Kumar - 7 years, 6 months ago

Can you explain.

Aejeth Lord - 7 years, 6 months ago

let,

$P(k)=\frac{k^{7}}{7}+\frac{k^{5}}{5}+\frac{2k^{3}}{3}-\frac{k}{105}$ is an integer for $k\in+Z$

$P(1)=\frac{1}{7}+\frac{1}{5}+\frac{2}{3}-\frac{1}{105}=1$ which is integer

$P(2)=\frac{2^7}{7}+\frac{2^5}{5}+\frac{2^4}{3}-\frac{2}{105}=30$ this is also integer

let, $P(n)=\frac{n^7}{7}+\frac{n^5}{5}+\frac{2n^3}{3}-\frac{n}{105}=N$ is an integer

where, $n>2$ and $n\in{+Z}$

Therefore $P(n+1)=\frac{(n+1)^7}{7}+\frac{(n+1)^5}{5}+\frac{2(n+1)^3}{3}-\frac{(n+1)}{105}$

$=(\frac{n^7}{7}+\frac{n^5}{5}+\frac{2n^3}{3}-\frac{n}{105})+(\frac{1}{7}+\frac{1}{5}+\frac{2}{3}-\frac{1}{105})+(n^6+3n^5+6n^4+7n^3+7n^2+4n)$

$=P(n)+P(1)+(n^6+3n^5+6n^4+7n^3+7n^2+4n)$

$=N+1+(n^6+3n^5+6n^4+7n^3+7n^2+4n)$

now, $P(1)$ is integer,we assume $P(n)=N$ is integer for some $n\in{+Z}$

and for any value of $n\in{+Z}$ , $(n^6+3n^5+6n^4+7n^3+7n^2+4n)$ is integer.

so, $P(n+1)$ is integer.

Thus,when ever the statement is true for $n$ it is true for $(n+1)$ .we have checked it is true for k=1,2.so by principle of induction it is true for all positive integer.

Gypsy Singer - 7 years, 6 months ago

@Gypsy Singer – Thank you very much.

Aejeth Lord - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$