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.

  1. If m,n,p,qm,n,p,q are non negative integers prove that, m=0q(nm)(p+m)!m!=(p+q+1)!q!(np+1qp+2)\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})

  2. Prove k77+k55+2k33k105 \dfrac{k^7}{7}+\dfrac{k^5}{5}+\dfrac{2k^3}{3}-\dfrac{k}{105} is an positive integer for all positive integers kk.

  3. If m>nm>n prove that a2n+1a^{2^ n}+1 is a divisor of a2m1a^{2^m}-1. Find LCM of (a2n+1,a2m1)(a^{2^n}+1, a^{2^m}-1).

Thanks in advance.

Note by Aejeth Lord
5 years, 10 months ago

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18 votes

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

敬全 钟 - 5 years, 10 months ago

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

Aejeth Lord - 5 years, 10 months ago

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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=1,0,1k = -1, 0, 1, and these are unique solutions. But I find that the question is a bit more complicated than this.

Michael Tong - 5 years, 10 months ago

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

The expression is equivalent to 1105(15k7+21k5+70k3k)\frac{1}{105}(15k^7+21k^5+70k^3-k), so if we can show that 15k7+21k5+70k3k15k^7+21k^5+70k^3-k is divisible by 105=357105=3*5*7 for all positive integral kk then we are done. Consider the expression modulo 33, 55, and 77. We have the equivalences k3k0(mod3)k^3-k \equiv 0 (mod 3), k5k0(mod5)k^5-k \equiv 0 (mod 5), and k7k0(mod7)k^7-k \equiv 0 (mod 7), which are all true for all positive integers kk by Fermat's little theorem, so the expression is divisible by 105105.

Logan Dymond - 5 years, 10 months ago

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

we can write,

(a2m1)=(a2m1+1)(a2m2+1)(a2m3+1)(a2+1)(a+1)(a1)(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=(mq) n=(m-q) where q1,2,3,,(m1) q\in{{1,2,3,\dots,(m-1)}}

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

so it is clear that for any value of q, (a2n+1) (a^{2{^n}}+1) is a factor of (a2m1) (a^{2^{m}}-1)

hence,(a2n+1)(a^{2^{n}}+1) is a divisor of (a2m1)(a^{2^{m}}-1) if m>n m>n

since, (a2n+1) (a^{2^{n}}+1) is a divisor of (a2m1) (a^{2^{m}}-1) so LCM of this two =(a2m1)(a^{2^{m}}-1)

Gypsy Singer - 5 years, 10 months ago

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the second question can be done using induction and binomial theroem

Siddharth Kumar - 5 years, 10 months ago

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Can you explain.

Aejeth Lord - 5 years, 10 months ago

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let,

P(k)=k77+k55+2k33k105 P(k)=\frac{k^{7}}{7}+\frac{k^{5}}{5}+\frac{2k^{3}}{3}-\frac{k}{105} is an integer for k+Zk\in+Z

P(1)=17+15+231105=1P(1)=\frac{1}{7}+\frac{1}{5}+\frac{2}{3}-\frac{1}{105}=1 which is integer

P(2)=277+255+2432105=30P(2)=\frac{2^7}{7}+\frac{2^5}{5}+\frac{2^4}{3}-\frac{2}{105}=30 this is also integer

let,P(n)=n77+n55+2n33n105=NP(n)=\frac{n^7}{7}+\frac{n^5}{5}+\frac{2n^3}{3}-\frac{n}{105}=N is an integer

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

Therefore P(n+1)=(n+1)77+(n+1)55+2(n+1)33(n+1)105P(n+1)=\frac{(n+1)^7}{7}+\frac{(n+1)^5}{5}+\frac{2(n+1)^3}{3}-\frac{(n+1)}{105}

=(n77+n55+2n33n105)+(17+15+231105)+(n6+3n5+6n4+7n3+7n2+4n) =(\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)+(n6+3n5+6n4+7n3+7n2+4n) =P(n)+P(1)+(n^6+3n^5+6n^4+7n^3+7n^2+4n)

=N+1+(n6+3n5+6n4+7n3+7n2+4n)=N+1+(n^6+3n^5+6n^4+7n^3+7n^2+4n)

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

and for any value of n+Zn\in{+Z} , (n6+3n5+6n4+7n3+7n2+4n) (n^6+3n^5+6n^4+7n^3+7n^2+4n) is integer.

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

Thus,when ever the statement is true for nn it is true for (n+1)(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 - 5 years, 10 months ago

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@Gypsy Singer Thank you very much.

Aejeth Lord - 5 years, 10 months ago

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