# 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 $$m,n,p,q$$ are non negative integers prove that, $$\displaystyle \sum_{m=0}^{q} (n-m)\dfrac {(p+m)!}{m!}= \dfrac {(p+q+1)!}{q!} \cdot (\dfrac {n}{p+1}-\dfrac{q}{p+2})$$

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 $$k$$.

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

Note by Aejeth Lord
4 years, 6 months ago

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

- 4 years, 6 months ago

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

- 4 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.

- 4 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 $$105$$.

- 4 years, 6 months ago

the second question can be done using induction and binomial theroem

- 4 years, 6 months ago

Can you explain.

- 4 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.

- 4 years, 6 months ago

Thank you very much.

- 4 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 =$$(a^{2^{m}}-1)$$

- 4 years, 6 months ago