Australian School of Excellence 2015 Number Theory Exam

1)If r is an odd prime, let it be of the form 2n+1 where n is a positive integer. If we put this value of r in our equation, we get 2(n+1)q-2np=(2n+1)^2. As L.H.S. is even and R.H.S is odd we won't have any solutions for p,q,r in this case. So r=2. We get 3q-p=4 which means q=(p+4)/3. We can see that q can be an integer when p=2,5,11,17,23,29,41,47,53,59,71,..... After which p+q will exceed 111. So when p is 71 then q is not prime, when p is 59 q is 21 which is again not a prime. But when p is 53, q is 19 which is indeed a prime. So our largest solution set for p,q is 53 and 19 and r=2. So largest possible value of pqr is 53×19×2=2014.

$\Large{{ 5 }^{ a }+{ b }^{ 2 }={ 3 }^{ c }\\ \Longrightarrow \frac { { 5 }^{ a }+{ b }^{ 2 } }{ 3 } ={ 3 }^{ c-1 }}$

Since $3^{c-1}$ is an integer. We have

${ 5 }^{ a }+{ b }^{ 2 }\equiv 0\quad \left( \mod 3 \right)$

If ${ b }^{ 2 }\equiv 0\quad \left( \mod 3 \right)$, So ${ 5 }^{ a }\equiv 0\quad \left( \mod 3 \right)$ (this is not possible). So by Fermat's Little theorem $b^2 \equiv 1 ( \mod 3) \Longrightarrow 5^a \equiv 2 \quad ( \mod 3)$.

This helps in creating a table for $5^a$:

\Large{\underline { \begin{matrix} a & & R \end{matrix} } \\ \begin{matrix} 1 & & 2 \\ 2 & & 1 \\ 3 & & 2 \\ 4 & & 1 \end{matrix} \\ \text{ Here R means the remainder when 5^a is divided by 3} }

Clearly $a$ is odd because at odd powers of 5 it leaves remainder 2.

$\Large{{ 5 }^{ a }+{ b }^{ 2 }={ 3 }^{ c }\\ \Longrightarrow { 5 }^{ a-1 }=\frac { { 3 }^{ c }-{ b }^{ 2 } }{ 5 } }$

If $b^2 \equiv 0 \quad (\mod 5)$, So $3^c \equiv 0 \quad ( \mod 5)$ (this is not possible). Hence we create a table for $3^c$:

$\Large{\underline { \begin{matrix} c & & R \end{matrix} } \\ \begin{matrix} 1 & & 3 \\ 2 & & 4 \\ 3 & & 2 \\ 4 & & 1 \end{matrix} }$

Clearly $c$ must be even because after only this $3^c \equiv b^2 \quad (\mod 5)$.

Say $a=3$ then we have $3^c > 125$, so minimum possible value of $c$ is $6$, but $729-125=604$ is not a perfect square.

By the same we can prove there will be no solutions for $a \ge 3$. This leaves only solution to us as $(a,b,c)=(1,2,2)$.

@Sharky Kesa

Here goes my solutions:

1)Pretty much the same as Kushagra. (though much longer)

4)Lemma 1: In the equation $5^a +b^2= 3^c$, $c$ is always even.
Proof: $3^c -b^2 \equiv 0 \mod{5}$
Five cases with $b \equiv 0-4 mod 5$
If, $b \equiv 1; c=4k,k \in \mathbb{Z}$
$b \equiv 2; c=4k+2,k \in \mathbb{Z}$
$b \equiv 3; c=4k+2,k \in \mathbb{Z}$
$b \equiv 4; c=4k,k \in \mathbb{Z}$
$b \equiv 1$ no solutions.
Hence $c$ is always even. And we are done.

Let $c=2m$, $5^a=3^c-b^2$
$5^a=3^{2m}-b^2$
$5^a=(3^m-b)(3^m +b)$
Hence we have two equations:
$5^l= 3^m -b$ $5^k=3^m +b; l+k=a$ Solving for $3^m$ and $b$, we get $b=\dfrac{5^k -5^l}{2}$ $3^m=\dfrac{5^k +5^l}{2} \longrightarrow (1)$ Now note that $l<k$ and also if $l,k>1$ then the equation on the right will not be a perfect power of 3, since we could factor out 5 from the equation. Thus we must have $l=0$ in (1).
We now have to solve this equation for $k,m$. [3^m =\dfrac{1 +5^k}{2} \longrightarrow (2)) for $k,m$

**Lemma 2: $k$ is odd in (2).**
We have
$1+5^k \equiv 2 \mod{3}$
$5^k \equiv 2 \mod{3}$ If $k$ is even we have $5^k \equiv 1 \mod{3}$ by Fermat's little theorem and so for $k$ odd $5^k\equiv 2 \mod{3}$ and we are done.

Lemma 3: There are no solutions to (2) if $k>3$.
Factoring some the terms, note that $k-1$ is even (and so $k-2$ odd).
$3^m= \dfrac{(5+1)(5^{k-1} -5^{k-2}+ 5^{k-3} \cdots +1)}{2}$
$3^{m-1}=(5^{k-1} -5^{k-2})+ (5^{k-3} -5^{k-4})+ \cdots +1$
Now see that each of the bracket $(5^{k-a}-5^{k-a-1}) \equiv -1 \mod{3}$. So you have a stream of (-1)'s followed by a (+1). So since
$(-1)+(-1)+(-1)+(-1) \cdots +1 \equiv 0 \mod{3}$
Thus, the number of (-1)'s are $\equiv 1 \mod{3}$
$k-1\equiv 2 \mod{3}$
$k\equiv 0 \mod 3$
Our second observation is that k is divisible by 3 along with being odd(Lemma 2). Now group the terms in triplesc(we can since there are k terms) and factor:
$3^{m-1}=5^{k-3}(5^2 -5 +1) -5^{k-6}(5^2 -5 +1)+ \cdots (5^2 -5 +1)$
$3^{m-1}=(5^2-5+1)(5^{k-3}-5^{k-6}+5^{k-9}-\cdots +1)$
$3^{m-1}=(21)(5^{k-3}-5^{k-6}+5^{k-9}-\cdots +1)$
This is a contradiction since The R.H.S is divisible by 7 and the L.H.S is not. And we are done! Q.E.D

Using Lemma 3 we have that (at last) $k=1, l=0 \implies a=1$. This means that $3^m =\dfrac{1 +5^1}{2} =3^1 \implies m=1 \implies c=2$ And finally $b=2$

Hence the only solution set to $5^a +b^2= 3^c$ is $(a,b,c) \equiv (1,2,2)$. And we are done.

Great Problem! (moving to the next set....)

You may find this helpful.

$\frac{16}{64} = \frac{1\cancel{6}}{\cancel{6}4}=\frac{1}{4}$

Answer to question no. 5 --> if d is the required no. And d=a^2+b^2-c^2 Then a,b,c is set of all triplets a,b,c where a,b,c are sides of an acute angled triangle and c is the largest side EXPLANATION- the given conditions can be manipulated as ●a^2+b^2>c^2 ●a<c ●b<c and this conditions are fullfilled by sides of an acute angled triangle. :D •smallest set of a,b,c is (4,5,6) Which comes just after (3,4,5)the smallest set of primitive pythagorean triplet 3^2+4^2=5^2 4^2+5^2> 6^6 now since we have value of a,b,c we can easily calculate values of d=a^2+b^2-c^2 Where d is the required numbers now for a,b,c(4,5,6) d=5 and so on

Hey sharky in q no.5 it wud b btr if u 'ill write a^2+b^2-c^2=d ,where d is some positive integer. without it question is not specific :)

Q3) Let the numbers be $AB=10a+b$ and $BC=10b+c$. Clearly we have

$\dfrac{10a+b}{10b+c}=\dfrac{a}{c}\\ \Longrightarrow b=\dfrac{9ac}{10a-c}$

Since $b \in \mathbb{Z^{+}}, 9 \ge b \ge 1$, We will make cases for $a=1,2,3,4,5,6,7,8,9$

For $a=1$

$b=\dfrac{9c}{10-c}$

By keeping the values of $c$ from $1$ to $9$, we get possible pairs as $(1,1,1), (1,6,4), (1,9,5)$. Applying same methods we find $8$ more tuples of $a=b=c$ and more solution as $(2,6,5), (4,9,8)$, therefore in total we have $\boxed{13}$ solutions.

If you feel my answer is wrong somewhere or you want to make this solution better please feel free to ask or reply me here.

Thank you @Sharky Kesa, since we are given numerator is strictly smaller than denominator so we discard the solutions of $a=b=c$. This gives only 4 possible solution as mentioned above.

Sharky I am getting only one solution to q4. Am I right? If I am I will post a solution soon