RMO 2015

5) We consider case when $p=2$. Clearly $2^2+3^2 = 4+9 = 13$ which is not a perfect square.So now consider $p>2$. Let if possible $2^p+3^q=q^2$ for some integer $q$.We note that $2^p \equiv 0 \pmod{4}$ , $3^p \equiv 3 \pmod{4}$ , thus $2^p+3^p \equiv 3 \pmod{4}$ but $q^2 \equiv 0,1 \pmod{4}$ which is a contradiction.

7) If $P(x)$ is a polynomial with integer coefficients then for distinct integers $a$, $b$, we have $a-b|P(a)-P(b)$.

So, it implies that $a-b|P(a)-P(b) = b-c$. Similarly we get, $b-c|c-a$ and $c-a|a-b$. So finally what we get is $a-b|b-c$, $b-c|c-a$ and $c-a | a-b$, this is possible iff $a-b = b-c =c-a$. But this implies that $a=b=c$. This is a contradiction as $a$, $b$ and $c$ are distinct. Therefore no such polynomial exists.

9) $x^3=x+1 \Rightarrow x^7=x^4(x+1) = x^5+x^4 = x^2(x+1)+x(x+1) = x^3+x^2+x^2+x = 2x^2+2x+1 \\ \Rightarrow (a,b,c)=(2,2,1)$

9)$x^3=x+1\Rightarrow x^6=(x^2+2x+1)\Rightarrow x^7=x^3+2x^2+x=2x^2+2x+1$.

For the $3rd$ question,

The sequence is, $S=1+3+6+10+15........+{t}_{n}$

To get the ${t}_{n}$ we can write it as,

$S=0+1+3+6+10+15........+{t}_{n}$

$S=1+3+6+10+15+.....$

Now, Subtracting both the sums,

$0=-1-2-3-4-5-.......{t}_{n}$

Therefore we can get,

${t}_{n}=\frac {n(n+1)}{2}$

Now to prove that there are infinitely many squares, Assume that ${t}_{n}$ is a square. From here we see that if ${t}_{n}$ is a square,

${t}_{4n(n+1)}=4\frac{n(n+1)}{2}.{(2n+1)}^{2}$ is also a square.

Hence if ${t}_{1}=1$ is a square then ${t}_{8}=36$ is a square. And if ${t}_{8}$ is a square then ${t}_{4*8(8+1)}={t}_{288}=144*289$ is a square.

Therefore,we get a sequence of infinite squares from here.

6) $a^{m}-b^{m}=(a-b)(a^{m-1}+a^{m-2}b+\ldots+ab^{m-2}+b^{m-1})$.

Since $a \equiv b \pmod {m^{n}},a \equiv b \pmod {m}$.

Hence, $a^{m-i}b^{i} \equiv b^{m} \pmod {m}$ and $a^{m-1}+a^{m-2}b+\ldots+ab^{m-2}+b^{m-1} \equiv mb^{m} \equiv 0 \pmod {m}$

Hence, $a^{m}-b^{m}$ is divisible by $m^{n+1}$.

Note: This solution is not original.

Credit: An Excursion in Mathematics.

would it help?

The following results are by Fermet Little Theorm:

$a^p = a modp$

$b^p = b modp$

then $a^p + b^p = a + b modp$

now $(a + b)^p = a + b modp$

so $(a + b)^p = a^p + b^p modp$

let us try 2nd question :

we can make following case and i am writing totient function as E,

Case 1 - when $m$ is prime then E(m) = m - 1, which is even.

Case 2 - when $m$ is even, clearly it would contain 2 so it even.

So, what are today's problems?

Thanks!

@Adarsh Kumar I believe that you can add a comment to the board, whenever you add a question. otherwise, it would be difficult to keep track.

You can add a comment like "Q3. is up" or "Next question!" so that people who are interested or have commented on this board get a notification, and they can check it out. Thanks.

@Adarsh Kumar In question 7, please mention that a,b and c are distinct. :)

6) $3n + 1 = a^2$

$3n = a^2 - 1$

$3n = (a + 1)(a - 1)$

so $a + 1 = 3 so a = 2$ OR

$a - 1 = 3 so a = 4$

then $n = 1 or n = 5$

so there are two n...

For the $2nd$ question, Case $1$: If there is atleast one odd prime factor of $m$,

Then by Euler's Totient function corolary, We can say that if prime $p$ divides a number $m$ then $p-1$ divides $\phi(m)$ . As $p$ is odd, $p-1$ is even, Therefore,$2|\phi(m)$.

Case $2$: If there is no odd prime factor of $m$,

Then $m$ can be written as $m={2}^{k}$ We can easily say that $\phi(m)={2}^{k}-{2}^{k-1}$ Hence,$2|\phi(m)$

All questions are done!

First problem can be solved easily by using binomial therom

@Calvin Lin Sir, is it possible for you to close this note since we already have a part 2 for this note.

can anybody solve the rmo 2008 Maharashtra region problem 5

For the $8th$ question,

$3n=(k-1)(k+1)$

By the Prime factorisation principle, There is a unique prime factorisation for every number.

From here we can say that,

$3=k-1$ and $k+1=n$

Therefore,$n=5$ is the only prime of this form.