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.

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TopNewestI 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 · 3 years, 8 months ago

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– Aejeth Lord · 3 years, 8 months ago

I am really sorry, I made a typo, I have corrected it now.Log in to reply

– Michael Tong · 3 years, 8 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.Log in to reply

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 · 3 years, 8 months ago

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the second question can be done using induction and binomial theroem – Siddharth Kumar · 3 years, 8 months ago

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– Aejeth Lord · 3 years, 8 months ago

Can you explain.Log in to reply

\( P(k)=\frac{k^{7}}{7}+\frac{k^{5}}{5}+\frac{2k^{3}}{3}-\frac{k}{105}\) 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 · 3 years, 8 months ago

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– Aejeth Lord · 3 years, 8 months ago

Thank you very much.Log in to reply

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 · 3 years, 8 months ago

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