Problems

Solution:

1. Suppose \(pqr = qrs\), since the roots are positive, we can divide by \(qr\) to obtain \(p=s\) (a contradiction. Therefore all those terms are distinct.
2. \(4c^3 = pqr+qrs+rsp+spq\), \(d^4 = pqrs\). Applying AM-GM, we obtain: \begin{align*} && c^3 = \frac{ pqr+qrs+rsp+spq}{4} & > \sqrt[4]{p^3q^3r^3s^3} = d^{3} \\ \Rightarrow && c &> d \end{align*}
3. There must be a turning point between each root (since there are no repeated roots).
4. \(f'(x) = 4x^3-12ax^2+12b^2-4c^3 = 4(x^3-3ax^2+3b^2-c^3)\). Letting the roots of this polynomial be \(\alpha, \beta, \gamma\) and again applying AM-GM, we must have: \begin{align*} && b^2 = \frac{\alpha\beta + \beta \gamma+\gamma \alpha}{3} &> \sqrt[3]{\alpha^2\beta^2\gamma^2} = c^2 \\ \Rightarrow && b &> c \end{align*}
5. Again, since there are turning points between the roots of \(f'(x)\) we must have distinct roots for \(f''(x)\), ie: \(f''(x) = 3x^2-6ax+6b^2 = 3(x^2-2ax+b^2)\) has distinct real roots. But for this to occur we must have that \((2a)^2-4b^2 = 4(a^2-b^2) > 0\), ie \(a>b\)

Solution:

1. \begin{align*} u = 1+(\alpha-1)x: && \int_0^1 (1 + (\alpha - 1)x)^n \d x &= \int_{u=1}^{u=\alpha} u^n \frac{1}{\alpha - 1} \d u \\ &&&= \left [\frac{u^{n+1}}{(n+1)(\alpha-1)} \right]_1^\alpha \\ &&&= \frac{\alpha^{n+1}-1}{(n+1)(\alpha-1)} \end{align*}
2. \begin{align*} && \int_0^1 (\alpha x + (1-x))^n \d x &= \int_0^1 \sum_{k=0}^n \binom{n}{k} \alpha^k x^k (1-x)^{n-k} \d x \\ &&&= \sum_{k=0}^n \alpha^k \int_0^1 \binom{n}{k} x^k (1-x)^{n-k} \d x \end{align*} Therefore the coefficient of \(\alpha^k\) is \(\displaystyle \int_0^1 \binom{n}{k} x^k (1-x)^{n-k} \d x\)
3. The coefficient of \(\alpha^{k}\) in \(\displaystyle \frac{\alpha^{n+1}-1}{(n+1)(\alpha-1)}\) is \(\displaystyle \frac1{n+1}\). Therefore \begin{align*} && \frac1{n+1} &= \binom{n}{k} \int_0^1 x^k(1-x)^{n-k} \d x \\ \Rightarrow && \int_0^1 x^k (1-x)^{n-k} \d x &= \frac{k!(n-k)!}{(n+1)n!} \\ &&&= \frac{k!(n-k)!}{(n+1)!} \end{align*}

Solution:

1. \(n^5 -n^3 = n^3(n-1)(n+1)\). If \(n\) is even then \(8 \mid n^3\). if \(n\) is odd then both of \(n-1\) and \(n+1\) are divisible by \(2\) and one is divisible by \(4\), so regardless \(8\) divides our expression. We can write \(n = 3k, 3k+1, 3k+2\) and in all cases our expression is divisible by \(3\). \(n = 3k \Rightarrow 3 \mid n\), \(n = 3k+1 \Rightarrow 3 \mid n-1\), \(n = 3k+2 \Rightarrow 3 \mid n+1\). Therefore \(3\) and \(8\) both divide our expression, and they are coprime so their product (24) divides our expression.
2. \(2^{2n}-1 = (2^2-1) \cdot (1+2^2+\cdots + 2^{2n-2}) = 3 \cdot N\) therefore \(3\) divides our number.
3. Suppose \(n-1 = 3k\) then \(n^3-1 = (3k+1)^3-1 = 27k^3 + 27k^2 + 9k\) which is clearly divisible by 9

Solution:

1. \(\,\) \begin{align*} && 0.383838\ldots &= \frac{3}{10} + \frac{8}{100} + \frac{3}{1000} + \cdots \\ &&&= \frac{38}{100} + \frac{38}{10000} + \cdots \\ &&&= \frac{38}{100} \left (1 + \frac{1}{100} + \frac{1}{100^2} + \cdots \right) \\ &&&= \frac{\frac{38}{100}}{1 - \frac{1}{100}} \\ &&&= \frac{38}{99} \end{align*}
2. Let \(x = 0\cdot a_{1}a_{2}a_{2}\cdots\) such that there exists \(N\), \(k\) such that \(a_{n+k} = a_n\) for \(n > N\). First notice that \begin{align*} x &= \frac{a_1a_2\cdots a_N}{10\ldots000} + \frac{a_{N+1}}{10^{N+1}} + \cdots \\ &= \frac{b}{10^N} + \frac{a_{N+1}a_{N+2}\cdots a_{N+k}}{10^{N+k}} + \frac{a_{N+1}a_{N+2}\cdots a_{N+2k}}{10^{N+k}} + \cdots \\ &= \frac{b}{10^N} + \frac{c}{10^{N+k}} \left (1 + \frac{1}{10^k} + \cdots \right) \\ &= \frac{b}{10^N} + \frac{c}{10^{N+k}} \frac{1}{1- \frac{1}{10^k}} \\ &= \frac{b}{10^N} + \frac{c}{10^{N+k}} \frac{10^k}{10^k-1} \\ &= \frac{b}{10^N} + \frac{c}{10^N(10^k-1)} \end{align*} where \(b = a_1a_2\cdots a_N\) and \(c = a_{N+1}a_{N+2}\cdots a_{N+k}\). Clearly as the sum of two rational numbers, \(x\) is rational.

Solution: \(12\) has factors \(1,2,3,4,6,12\) of which \(4\) are neither \(1\) nor \(12\). \(16\) has factors \(1,2,4,8,16\) of which \(3\) are neither \(1\) nor \(16\). If \(N = p_1^{m_1} \cdots p_r^{m_r}\) then \(N\) has \((m_1+1)\cdots(m_r+1)\) factors since we can have between \(0 \leq k \leq m_i\) of the \(i\)th prime factor, whcih is \(m_i+1\) possibilities. We then need to subtract two for the proper factors, ie \((m_1+1)\cdots(m_r+1) - 2\).

1. We need \((m_1 + 1) \cdots (m_r + 1) - 2 = 12\) ie \((m_1+1) \cdots (m_r +1) = 14\). Since \(14 = 2 \times 7\) we should have two prime factors, ie of the form \(p_1^6p_2^1\). The smallest number of this form is \(2^6 \cdot 3 = 192\)
2. We wish to make \((m_1+1)\cdots(m_r+1) \geq 14\) with \(2^{m_1} \cdot 3^{m_2} \cdots p_r^{m_r}\) as small as possible and \(m_1 \geq m_2 \geq \cdots \geq m_r\). With \(1\) prime, the best we can do is \(2^{13}\). With \(2\) primes the best we can do is \(2^p \cdot 3^q\) with \((p+1)(q+1) \geq 14\). \begin{array}{c|c} q & p & N \\ \hline 0 & 13 & 2^{13} \\ 1 & 6 & 2^6 \cdot 3^1 \\ 2 & 5 & 2^5 \cdot 3^2 \\ 3 & 3.& 2^3 \cdot 3^3 \end{array} So the best here is \(2^6 \cdot 3\) With \(3\) primes, the best we can do is \(2^p 3^q 5^r\) with \(p \geq q \geq r\) and \((p+1)(q+1)(r+1) \geq 14\) \begin{array}{c|c|c|c} r & q & p & N \\ \hline 1 & 1 & 3 & 2^3 \cdot 3 \cdot 5 \\ 1 & 2 & 2& 2^2 \cdot 3^2 \cdot 5 \\ 2 & 2& 2 & 2^2 \cdot 3^2 \cdot 5^2 \end{array} With four primes we can achieve it immediately with \(2\cdot3\cdot5\cdot 7\) and more primes clearly wont help, so our best options is \(120\)

Solution: \begin{align*} && k &\leq 2(k-2) \\ \Rightarrow && 4 &\leq k \end{align*} Lemma: Suppose we construct \(N \neq \) (optimally) as a sum out of \(a_1 + \cdots +a_k\), then \(a_i \in \{2, 3\}\). Proof: Suppose not, suppose some \(a_i > 3\). Then from our earlier inequality, the sum \(a_1 + \cdots +a_{i-1} + 2 + (a_i - 2) + \cdots \) has the same sum, but a larger product. Therefore \(a_i \leq 3\). Suppose also some \(a_i = 1\), then we could replace \(a_1\) with \(a_1+1\) and remove \(a_i\), leaving us again with the same sum but larger product. (Assuming \(N \neq 1\)) \(5 = 2+3\) is the only way to write \(5\) as a sum of \(2\)s and \(3\)s, therefore \(P(5) = 2\times 3\) \(6 = 2 + 2 + 2 = 3 + 3\) and we can immediately see that \(2^3 = 8 < 3^2 = 9\), so \(P(6) = 3^2\) and whenever we have three \(2\)s we should replace them with two \(3\)s. So \(7 = 2 + 2 + 3 \Rightarrow P(7) = 2^2 \times 3\) \(8 = 3 + 3 + 2 \Rightarrow P(8) = 2\times 3^2\) \(9 = 3 + 3 + 3 \Rightarrow P(9) = 3^3\) Suppose \(1000 = 2n + 3m\), considered \(\pmod{3}\) we can see that \(n \equiv 2 \pmod{3}\) therefore we should have \(1000 = 2 + 2 + \underbrace{3 + \cdots + 3}_{332\text{ }3\text{s}}\) and so \(P(1000) = 2^2 \times 3^{332}\)

Solution:

1. Let $\mathbf{A} = \begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix}\(, then we need to show that \)(a\mathbf{A})(b\mathbf{A})\( is of the form \)cA\( where \)a, b, c \neq 0$. Since $\mathbf{A}^2 = \begin{pmatrix}2 & 2\\ 2 & 2 \end{pmatrix} = 2\mathbf{A}\( this is certainly the case, since \)(a\mathbf{A})(b\mathbf{A}) = 2ab\mathbf{A}$. To check that we have a group be need to check:
   • Closure (done)
   • Associativity (inherited from matrix multiplication)
   • Identity (\(\frac12 \mathbf{A}\))
   • Inverses the inverse of \(a\mathbf{A}\) is \(\frac{1}{4a}\mathbf{A}\)
2. Suppose \(\mathbf{A}\) is singular (ie \(\det\mathbf{A}=0\)), then \(\mathbf{AA^{-1}B=B}\) (where inverse is the group inverse rather than the matrix inverse) for any matrix \(\mathbf{B}\). Taking determinants we have: \(\det(\mathbf{AA^{-1}B}) = \det(B) \Rightarrow \det(A) \det(A^{-1}B) = \det(B) \Rightarrow 0 = \det(B)\), ie all matrices are singular. Consider the set of non-zero multiples of \(\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}\), then the same logic as part (i) will suffice

Solution: Definition: \(\displaystyle \mathbb{E}(X) = \sum_{i=1}^n x_i \mathbb{P}(X = x_i)\) Definition: \(\displaystyle \mathrm{Var}(X) = \sum_{i=1}^n (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i)\) Claim: \(\mathbb{E}(aX+b) = a\mathbb{E}(X)+b\) Proof: \begin{align*} \mathbb{E}(aX+b) &= \sum_{i=1}^n (ax_i+b) \mathbb{P}(X = x_i) \\ &= a\sum_{i=1}^n x_i \mathbb{P}(X = x_i) + b\sum_{i=1}^n \mathbb{P}(X = x_i)\\ &= a \mathbb{E}(X) + b \end{align*} Claim: \(\mathrm{Var}(aX+b) = a^2 \mathrm{Var}(X)\) Claim: \(\mathrm{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right)\leqslant\frac{\mathrm{var}(X)}{\lambda^{2}}\) Proof: \begin{align*} \mathrm{Var}(X) &= \sum_{i=1}^n (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) \\ &\geq \sum_{|x_i - \mathbb{E}(X)| \geq \lambda} (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) \\ &\geq \sum_{|x_i - \mathbb{E}(X)| \geq \lambda} \lambda^2 \mathbb{P}(X = x_i) \\ &= \lambda^2 \sum_{|x_i - \mathbb{E}(X)| \geq \lambda} \mathbb{P}(X = x_i) \\ &= \lambda^2 \mathrm{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right) \end{align*} Claim: \[ \mathrm{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right)\geqslant\frac{\mathrm{var}(X)-\lambda^{2}}{k^{2}-\lambda^{2}}\,. \] Proof: \begin{align*} && \mathrm{Var}(X) &= \sum_{i=1}^n (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) \\ &&&= \sum_{|x_i - \mathbb{E}(X)| \geq \lambda} (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) + \sum_{|x_i - \mathbb{E}(X)| < \lambda} (x_i-\mathbb{E}(X))^2 \mathbb{P}(X = x_i) \\ &&& \leq \sum_{|x_i - \mathbb{E}(X)| \geq \lambda} k^2 \mathbb{P}(X = x_i) + \sum_{|x_i - \mathbb{E}(X)| < \lambda} \lambda^2 \mathbb{P}(X = x_i) \\ &&&= k^2 \mathbb{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right) + \lambda^2 \mathbb{P}\left(\left|X-\mathrm{E}(X)\right| < \lambda\right) \\ &&&= k^2 \mathbb{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right) + \lambda^2(1- \mathbb{P}\left(\left|X-\mathrm{E}(X)\right| \leq \lambda\right) \\ &&&= (k^2 - \lambda^2) \mathbb{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right) + \lambda^2 \\ \Rightarrow&& \frac{\mathrm{Var}(X)-\lambda^2}{k^2 - \lambda^2} &\leq \mathbb{P}\left(\left|X-\mathrm{E}(X)\right|\geqslant\lambda\right) \end{align*} [Note: This result is known as Chebyshev's inequality, and is an important starting point to understanding the behaviour of tails of random variables]

Solution: Since \(A\) scored \(24\) points, he must have finished first in all but one race and second in that race. Given \(E\) won the final stage, \(A\) must have been \(11112\) \begin{array}{c|ccccc|c} & 1 & 2 & 3 & 4& 5 & \sum \\ \hline A & 1 & 1 & 1 & 1 & 2 & 24 \\ B & - & - & - & - & - & \\ C & - & - & - & - & - & \\ D & - & - & - & - & - & \\ E & - & - & - & 3 & 1 \\ \end{array} If \(E\) has \(12\) points the smallest number of points the others can have are \(13, 14, 15\) which would be a total of \(78\) points \(\geq 15 \times 5 = 75\), more than is available, therefore \(E\) must have the minimum \(11\) points. \begin{array}{c|ccccc|c} & 1 & 2 & 3 & 4& 5 & \sum \\ \hline A & 1 & 1 & 1 & 1 & 2 & 24 \\ B & - & - & - & - & - & \\ C & - & - & - & - & - & \\ D & - & - & - & - & - & \\ E & 5 & 5 & 5 & 3 & 1 & 11 \\ \end{array} There are now \(40\) points to be divided between \(B, C\) and \(D\). \(12+13+14 = 39\), so only way to achieve this is \(12, 13, 15\). \begin{array}{c|ccccc|c} & 1 & 2 & 3 & 4& 5 & \sum \\ \hline A & 1 & 1 & 1 & 1 & 2 & 24 \\ B & - & - & - & - & - & 15 \\ C & - & - & - & - & - & 13 \\ D & - & - & - & - & - & 12 \\ E & 5 & 5 & 5 & 3 & 1 & 11 \\ \end{array} Camus gained the same position in four of the five races. So we need \(4x + y = 13\) which can be done with \(4 \times 1 + 9\) or \(4 \times 2 + 5\) or \(4 \times 3 + 1\). The first two aren't possible (you can't score \(9\)) and the second isn't possible (all the first places are taken) so \(C\) must have four third places and a last place. (Which also must be the second to last race since there are alread last places in \(3\) of the races and a third place in the second to last) \begin{array}{c|ccccc|c} & 1 & 2 & 3 & 4& 5 & \sum \\ \hline A & 1 & 1 & 1 & 1 & 2 & 24 \\ B & - & - & - & - & - & 15 \\ C & 3 & 3 & 3 & 5 & 3 & 13 \\ D & - & - & - & - & - & 12 \\ E & 5 & 5 & 5 & 3 & 1 & 11 \\ \end{array} There are now one \(5\), five \(4\)s, four \(2\)s left to place. And they need to add to \(12\) for one rider. [In score terms this is \(1, 5\times 2, 4 \times 4\). Neither rider can have all the second places, and since they would score too highly, and \(D\) can't have more than one second place since otherwise he'd score too highly. Therefore \(B\) has three second places. So \(B\) is \(1,2,4,4,4\) and \(C\) is \(4,2,2,2,2\) in some order. \(D\) can't come second in the last race, so he comes \(4\)th and \(B\) comes \(5\)th \begin{array}{c|ccccc|c} & 1 & 2 & 3 & 4& 5 & \sum \\ \hline A & 1 & 1 & 1 & 1 & 2 & 24 \\ B & - & - & - & - & 5 & 15 \\ C & 3 & 3 & 3 & 5 & 3 & 13 \\ D & - & - & - & - & 4 & 12 \\ E & 5 & 5 & 5 & 3 & 1 & 11 \\ \end{array}

Solution:

1. \([AIB] = \frac12br\), \([BIC] = \frac12ar\), \([CIA] = \frac12 rc\), therefore \(\Delta = [AIB] +[BIC] + [CIA] = \frac12r(a+b+c) = sr\)
2. \(\,\) \begin{align*} && \Delta &= \frac12 bc \sin \alpha \\ \Rightarrow && \Delta^2 &= \frac14 b^2c^2 \sin^2 \alpha \\ &&&= \frac14 \left (b^2c^2 - b^2c^2\cos^2 \alpha \right) \\ &&&= \frac1{16} \left (4b^2c^2 - (2bc\cos \alpha )^2\right) \\ \\ \Rightarrow && \Delta^2 &= \frac1{16} \left (4b^2c^2 - (b^2+c^2-a^2 )^2\right) \\ &&&= \frac1{16} (2bc-b^2-c^2+a^2)(2bc+b^2+c^2-a^2) \\ &&&= \frac{1}{16}(a^2-(b-c)^2)((b+c)^2-a^2) \\ &&&= \frac1{16}(a-b+c)(a+b-c)(b+c-a)(b+c+a) \\ &&&= (s - b)(s-c)(s-a)s \\ \Rightarrow && \Delta &= \sqrt{s(s-a)(s-b)(s-c)} \end{align*}
3. We have the setting like this,
   

   + - ↺
   so \begin{align*} && h & = \sqrt{R^2-r^2} \\ &&&= \sqrt{R^2-\frac{\Delta^2}{s^2}} \\ &&&= \sqrt{R^2 - \frac{(s-a)(s-b)(s-c)}{s}} \end{align*}

Solution: \begin{array}{c|c} \text{ends in} & \text{multiply by} \\ \hline 1 & 1 \\ 3 & 7 \\ 7 & 3 \\ 9 & 9 \end{array} If if \(n = 2^a \cdot 5^b \cdot c\) where \(c\) has no factors of \(2\) and \(5\) then we can multiply by \(2^b \cdot 5^a\) to obtain \(c\) followed by \(0\)s. Since \(c\) is neither even, nor a multiple of \(5\), by the earlier part of the question we can find a multiple such that it ends in \(1\). Suppose it is a \(k\) digit number, the consider Now consider \(1\underbrace{00\cdots0}_{k\text{ digits}}2\underbrace{00\cdots0}_{k\text{ digits}}\cdots 8\underbrace{00\cdots0}_{k\text{ digits}}9\cdot 0\), then clearly each section will end in the leading digit (ie all digits from \(1\) to \(9\)) and also end with a \(0\)

Solution: \begin{align*} && P(x) &= (x-a)^mQ(x) \\ \Rightarrow && P'(x) &= m(x-a)^{m-1}Q(x) + (x-a)^mQ'(x) \\ &&&= (x-a)^{m-1}(\underbrace{mQ(x) + (x-a)Q'(x)}_{\text{a polynomial}}) \\ &&&= (x-a)^{m-1}R(x) \end{align*} Therefore \(P^{(r)}(a) = 0\) for \(1 \leq r \leq m-1\) since each time we differentiate we will have a factor of \((x-a)^{m-r}\) which is zero when we evaluate at \(x = a\). If \(P_n(x) = \frac{\d^n}{\d x^n}(x^2-1)^n\) then we are differentiating a degree \(2n\) polynomial \(n\) times. Each time we differentiate we reduce the degree by \(1\), therefore the degree of \(P_n\) is \(n\). \begin{align*} && \int_{-1}^1 x^mP_n(x) \d x &= \left [x^m \underbrace{\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right)}_{\text{has a factor of }x-1\text{ and }x+1}\right]_{-1}^1 - \int_{-1}^1 mx^{m-1}\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= 0 - \int_{-1}^1 mx^{m-1}\frac{\d^{n-1}}{\d x^{n-1}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= -\left [mx^{m-1} \underbrace{\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right)}_{\text{has a factor of }x-1\text{ and }x+1}\right]_{-1}^1+ \int_{-1}^1 m(m-1)x^{m-2}\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= m(m-1)\int_{-1}^1 x^{m-2}\frac{\d^{n-2}}{\d x^{n-2}}\left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&& \cdots \\ &&&= (-1)^m m!\int_{-1}^1 \frac{\d^{n-m}}{\d x^{n-m}} \left ( (x-1)^{n} (x+1)^{n} \right) \d x\\ &&&= 0 \end{align*} If \(n = m\), we have \begin{align*} && \int_{-1}^1 x^n P_n(x) \d x&= (-1)^nn! \int_{-1}^1 (x^2-1)^n \d x \\ && &= (-1)^{2n}n! \cdot 2\int_{0}^1 (1-x^2)^n \d x \\ x = \sin \theta, \d x = \cos \theta \d \theta: &&&= 2 \cdot n!\int_{0}^{\pi/2} \cos^{2n} \theta \cdot \cos \theta \d \theta \\ &&&= 2 \cdot n!\int_{0}^{\pi/2} \cos^{2n+1} \theta \d \theta \\ &&&= 2 \cdot n!\frac{(2^{2n})(n!)^{2}}{(2n+1)!} \\ &&&= \frac{(2^{2n+1})(n!)^{3}}{(2n+1)!} \\ \end{align*}

Solution: Let \(P(x,y)\) be the statement that the functional equation holds, then: \begin{align*} P(0,0): && f(0) &= f(0)f(0)-f(a)f(a) \\ \Rightarrow && 1 &= 1 - f(a)^2 \\ \Rightarrow && f(a)^2 &= 0 \\ \Rightarrow && f(a) &= 0 \end{align*}

1. \begin{align*} P(0,t): && f(-t) &= f(0)f(t) - f(a)f(a-t) \\ \Rightarrow && f(-t) &= f(t) - 0 \\ \Rightarrow && f(t) &= f(-t) \end{align*}
2. \begin{align*} P(a,a): && f(0) &= f(a)f(a)-f(0)f(2a) \\ \Rightarrow && 1 &= 0 - f(2a) \\ \Rightarrow && f(2a) &= -1 \end{align*}
3. \begin{align*} P(2a,t): && f(2a-t) &= f(2a)f(t) - f(-a)f(a+t) \\ \Rightarrow && f(2a-t) &= -f(t)-f(a)f(a+t) \\ &&&= -f(t)-0 \\ \Rightarrow && f(2a-t) &= -f(t) \end{align*}
4. \begin{align*} && f(4a+t) &= f(2a-(-2a-t)) \\ &&&=-f(2a+t) \\ &&&=-f(2a-(-t)) \\ &&&=f(-t) \\ &&&=f(t) \end{align*}

Solution:

1. Suppose \(p > 3\) then there are clearly no solutions, since \(\frac1p+\frac1q+\frac1r \leq \frac{1}{4} + \frac{1}{4} + \frac{1}{4} < 1\) Therefore there are 3 cases: \(p = 3 \Rightarrow p = q = r = 3\) \(p = 2\): \begin{align*} && \frac12 = \frac1q + \frac1r \\ \Rightarrow && 0 = qr - 2q-2q \\ \Rightarrow && 4 &= (q-2)(r-2) \\ \end{align*} Therefore \((p,q,r) = (2, 3, 6), (2, 4, 4)\) \(p = 1\) we have a contradiction the other way.
2. We have already shown \(p < 3\), so we just need to check \(p = 2\) (since \(p=1\) is described in the question). \begin{align*} && \frac12 &< \frac1q+\frac1r \\ \Rightarrow && qr &< 2q+2r \\ \Rightarrow && 4 &> (q-2)(r-2) \\ \end{align*} Therefore we can have \((q-2)(r-2) = 0 \Rightarrow p = 2, q = 2, r = n\) Or we have have \((q-2)(r-2) = 1 \Rightarrow q = 3, r = 3\) Or we can have \((q-2)(r-2) = 2 \Rightarrow q = 3, r= 4\)