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2024 Paper 2 Q1
D: 1500.0 B: 1500.0

In the equality \[ 4 + 5 + 6 + 7 + 8 = 9 + 10 + 11, \] the sum of the five consecutive integers from 4 upwards is equal to the sum of the next three consecutive integers. Throughout this question, the variables \(n\), \(k\) and \(c\) represent positive integers.

  1. Show that the sum of the \(n + k\) consecutive integers from \(c\) upwards is equal to the sum of the next \(n\) consecutive integers if and only if \[ 2n^2 + k = 2ck + k^2. \]
  2. Find the set of possible values of \(n\), and the corresponding values of \(c\), in each of the cases
    1. \(k = 1\)
    2. \(k = 2\).
  3. Show that there are no solutions for \(c\) and \(n\) if \(k = 4\).
  4. Consider now the case where \(c = 1\).
    1. Find two possible values of \(k\) and the corresponding values of \(n\).
    2. Show, given a possible value \(N\) of \(n\), and the corresponding value \(K\) of \(k\), that \[ N' = 3N + 2K + 1 \] will also be a possible value of \(n\), with \[ K' = 4N + 3K + 1 \] as the corresponding value of \(k\).
    3. Find two further possible values of \(k\) and the corresponding values of \(n\).


Solution:

  1. Suppose the sum of the \(n + k\) consecutive integers from \(c\) upwards is equal to the sum of the next \(n\) consecutive integers then \begin{align*} && \sum_{i=c}^{i=c+n+k-1} i &= \sum_{i=c+n+k}^{c+2n+k-1} i \\ \Leftrightarrow && \frac{(c+n+k-1)(c+n+k)}{2} - \frac{(c-1)c}{2} &= \frac{(c+2n+k-1)(c+2n+k)}{2} - \frac{(c+n+k-1)(c+n+k)}{2} \\ \Leftrightarrow && 2(c+n+k-1)(c+n+k) &= (c+2n+k-1)(c+2n+k) + c(c-1) \\ \Leftrightarrow && 2c^2+4cn+4ck+2n^2+4kn+2k^2-2c-2n-2k&=2c^2+4cn+2ck+4n^2+4nk+k^2-2c-2n-k \\ \Leftrightarrow && 2ck+k^2&=2n^2+k \\ \end{align*}
    1. If \(k=1\) then \begin{align*} && 2n^2 + 1 &= 2c + 1 \\ \Rightarrow && c &= n^2 \end{align*} So \(n\) can take any value and \(c = n^2\)
    2. If \(k=2\) then \begin{align*} && 2n^2+2&= 4c+4 \\ \Rightarrow && n^2-1 &=2c \end{align*} So \(n\) must be odd, and \(c = \frac12(n^2-1)\)
  2. Suppose \(k=4\) then \(2n^2+4 = 8c+16\) or \(n^2-6 = 4c\) but then the left hand side is \(2, 3 \pmod{4}\) which is a contradiction.
  3. Suppose \(c =1\)
    1. Since \(2n^2+k = 2k + k^2\) or \(2n^2 = k^2+k\) we can have \(k = 1, n = 1\) or \(k = 8, n = 6\)
    2. Suppose \(2N^2 = K^2 + K\) then consider \begin{align*} && 2(N')^2 &= 2(3N+2K+1)^2 \\ &&&= 2(9N^2+4K^2+1+12NK+6N+4K) \\ &&&= 18N^2+8K^2+24NK+12N+8K+2 \\ && (K')^2+K' &= (4N+3K+1)^2 + (4N+3K+1) \\ &&&= 16N^2 + 9K^2+1+24NK+12N+9K+1 \\ &&&= 16N^2+9K^2+24NK+12N+9K+2 \\ \Rightarrow && 2(N')^2-(K')^2-K' &= 2N^2-K^2-K \\ &&&= 0 \end{align*} as required.
    3. So consider \((k,n) = (1,1), (8,6), (49, 35), (288,204)\)

2023 Paper 2 Q7
D: 1500.0 B: 1500.0

  1. The complex numbers \(z\) and \(w\) have real and imaginary parts given by \(z = a + \mathrm{i}b\) and \(w = c + \mathrm{i}d\). Prove that \(|zw| = |z||w|\).
  2. By considering the complex numbers \(2 + \mathrm{i}\) and \(10 + 11\mathrm{i}\), find positive integers \(h\) and \(k\) such that \(h^2 + k^2 = 5 \times 221\).
  3. Find positive integers \(m\) and \(n\) such that \(m^2 + n^2 = 8045\).
  4. You are given that \(102^2 + 201^2 = 50805\). Find positive integers \(p\) and \(q\) such that \(p^2 + q^2 = 36 \times 50805\).
  5. Find three distinct pairs of positive integers \(r\) and \(s\) such that \(r^2 + s^2 = 25 \times 1002082\) and \(r < s\).
  6. You are given that \(109 \times 9193 = 1002037\). Find positive integers \(t\) and \(u\) such that \(t^2 + u^2 = 9193\).

2023 Paper 3 Q5
D: 1500.0 B: 1500.0

  1. Show that if \[\frac{1}{x} + \frac{2}{y} = \frac{2}{7}\,,\] then \((2x - 7)(y - 7) = 49\). By considering the factors of \(49\), find all the pairs of positive integers \(x\) and \(y\) such that \[\frac{1}{x} + \frac{2}{y} = \frac{2}{7}\,.\]
  2. Let \(p\) and \(q\) be prime numbers such that \[p^2 + pq + q^2 = n^2\] where \(n\) is a positive integer. Show that \[(p + q + n)(p + q - n) = pq\] and hence explain why \(p + q = n + 1\). Hence find the possible values of \(p\) and \(q\).
  3. Let \(p\) and \(q\) be positive and \[p^3 + q^3 + 3pq^2 = n^3\,.\] Show that \(p + q - n < p\) and \(p + q - n < q\). Show that there are no prime numbers \(p\) and \(q\) such that \(p^3 + q^3 + 3pq^2\) is the cube of an integer.

2022 Paper 3 Q2
D: 1500.0 B: 1500.0

  1. Suppose that there are three non-zero integers \(a\), \(b\) and \(c\) for which \(a^3 + 2b^3 + 4c^3 = 0\). Explain why there must exist an integer \(p\), with \(|p| < |a|\), such that \(4p^3 + b^3 + 2c^3 = 0\), and show further that there must exist integers \(p\), \(q\) and \(r\), with \(|p| < |a|\), \(|q| < |b|\) and \(|r| < |c|\), such that \(p^3 + 2q^3 + 4r^3 = 0\). Deduce that no such integers \(a\), \(b\) and \(c\) can exist.
  2. Prove that there are no non-zero integers \(a\), \(b\) and \(c\) for which \(9a^3 + 10b^3 + 6c^3 = 0\).
  3. By considering the expression \((3n \pm 1)^2\), prove that, unless an integer is a multiple of three, its square is one more than a multiple of \(3\). Deduce that the sum of the squares of two integers can only be a multiple of three if each of the integers is a multiple of three. Hence prove that there are no non-zero integers \(a\), \(b\) and \(c\) for which \(a^2 + b^2 = 3c^2\).
  4. Prove that there are no non-zero integers \(a\), \(b\) and \(c\) for which \(a^2 + b^2 + c^2 = 4abc\).

2022 Paper 3 Q8
D: 1500.0 B: 1500.0

  1. Use De Moivre's theorem to prove that for any positive integer \(k > 1\), \[ \sin(k\theta) = \sin\theta\cos^{k-1}\theta \left( k - \binom{k}{3}(\sec^2\theta - 1) + \binom{k}{5}(\sec^2\theta - 1)^2 - \cdots \right) \] and find a similar expression for \(\cos(k\theta)\).
  2. Let \(\theta = \cos^{-1}(\frac{1}{a})\), where \(\theta\) is measured in degrees, and \(a\) is an odd integer greater than \(1\). Suppose that there is a positive integer \(k\) such that \(\sin(k\theta) = 0\) and \(\sin(m\theta) \neq 0\) for all integers \(m\) with \(0 < m < k\). Show that it would be necessary to have \(k\) even and \(\cos(\frac{1}{2}k\theta) = 0\). Deduce that \(\theta\) is irrational.
  3. Show that if \(\phi = \cot^{-1}(\frac{1}{b})\), where \(\phi\) is measured in degrees, and \(b\) is an even integer greater than \(1\), then \(\phi\) is irrational.

2020 Paper 2 Q5
D: 1500.0 B: 1500.0

If \(x\) is a positive integer, the value of the function \(\mathrm{d}(x)\) is the sum of the digits of \(x\) in base 10. For example, \(\mathrm{d}(249) = 2 + 4 + 9 = 15\). An \(n\)-digit positive integer \(x\) is written in the form \(\displaystyle\sum_{r=0}^{n-1} a_r \times 10^r\), where \(0 \leqslant a_r \leqslant 9\) for all \(0 \leqslant r \leqslant n-1\) and \(a_{n-1} > 0\).

  1. Prove that \(x - \mathrm{d}(x)\) is non-negative and divisible by \(9\).
  2. Prove that \(x - 44\mathrm{d}(x)\) is a multiple of \(9\) if and only if \(x\) is a multiple of \(9\). Suppose that \(x = 44\mathrm{d}(x)\). Show that if \(x\) has \(n\) digits, then \(x \leqslant 396n\) and \(x \geqslant 10^{n-1}\), and hence that \(n \leqslant 4\). Find a value of \(x\) for which \(x = 44\mathrm{d}(x)\). Show that there are no further values of \(x\) satisfying this equation.
  3. Find a value of \(x\) for which \(x = 107\mathrm{d}\left(\mathrm{d}(x)\right)\). Show that there are no further values of \(x\) satisfying this equation.

2020 Paper 3 Q8
D: 1500.0 B: 1500.0

A sequence \(u_k\), for integer \(k \geqslant 1\), is defined as follows. \[ u_1 = 1 \] \[ u_{2k} = u_k \text{ for } k \geqslant 1 \] \[ u_{2k+1} = u_k + u_{k+1} \text{ for } k \geqslant 1 \]

  1. Show that, for every pair of consecutive terms of this sequence, except the first pair, the term with odd subscript is larger than the term with even subscript.
  2. Suppose that two consecutive terms in this sequence have a common factor greater than one. Show that there are then two consecutive terms earlier in the sequence which have the same common factor. Deduce that any two consecutive terms in this sequence are co-prime (do not have a common factor greater than one).
  3. Prove that it is not possible for two positive integers to appear consecutively in the same order in two different places in the sequence.
  4. Suppose that \(a\) and \(b\) are two co-prime positive integers which do not occur consecutively in the sequence with \(b\) following \(a\). If \(a > b\), show that \(a-b\) and \(b\) are two co-prime positive integers which do not occur consecutively in the sequence with \(b\) following \(a-b\), and whose sum is smaller than \(a+b\). Find a similar result for \(a < b\).
  5. For each integer \(n \geqslant 1\), define the function \(\mathrm{f}\) from the positive integers to the positive rational numbers by \(\mathrm{f}(n) = \dfrac{u_n}{u_{n+1}}\). Show that the range of \(\mathrm{f}\) is all the positive rational numbers, and that \(\mathrm{f}\) has an inverse.

2019 Paper 1 Q7
D: 1500.0 B: 1500.0

Consider the following steps in a proof that \(\sqrt{2} + \sqrt{3}\) is irrational.

  1. If an integer \(a\) is not divisible by 3, then \(a = 3k \pm 1\), for some integer \(k\). In both cases, \(a^2\) is one more than a multiple of 3.
  2. Suppose that \(\sqrt{2} + \sqrt{3}\) is rational, and equal to \(\frac{a}{b}\), where \(a\) and \(b\) are positive integers with no common factor greater than one.
  3. Then \(a^4 + b^4 = 10a^2b^2\).
  4. So if \(a\) is divisible by 3, then \(b\) is divisible by 3.
  5. Hence \(\sqrt{2} + \sqrt{3}\) is irrational.
  1. Show clearly that steps 1, 3 and 4 are all valid and that the conclusion 5 follows from the previous steps of the argument.
  2. Prove, by means of a similar method but using divisibility by 5 instead of 3, that \(\sqrt{6} + \sqrt{7}\) is irrational. Why can divisibility by 3 not be used in this case?


Solution:

  1. Step 1: There are only three possibilities for the remainder of \(a\) when divided by \(3\), (\(0\), \(1\), \(2\)). \(a = 3m+r\). If \(r = 0\) we are done, if \(r = 1\) take \(k = m\), and \(r=2\) take \(k=(m+1)\) and we have \(a = 3k-1\) as required. Then \(a^2 = (3k\pm1)^2 =9k^2\pm6k+1 = 3(3k^32\pm2k)+1\) which is clearly \(1\) more than a square. Step 3: \begin{align*} && \frac{a}{b} &= \sqrt{2}+\sqrt{3} \\ \Rightarrow && \frac{a^2}{b^2} &= 5+2\sqrt{6} \\ \Rightarrow && \frac{a^2}{b^2}-5 &= 2\sqrt{6} \\ \Rightarrow && 24 &= \left ( \frac{a^2}{b^2}-5 \right)^2 \\ &&&= 25 + \frac{a^4}{b^4}-10\frac{a^2}{b^2} \\ \Rightarrow && -b^4 &= a^4-10a^2b^2 \\ \Rightarrow && a^4+b^4 &= 10a^2b^2 \end{align*} Step 4: If \(a\) is divisible by \(3\) then \(b^4 = 10a^2b^2-a^4\) is a multiple of \(3\), but if \(b\) was not a multiple of \(3\) then \(b^2\) would be \(1\) more than a multiple of \(3\) (by Step 3) and \(b^4\) would also be \(1\) more than a multiple of \(3\), and we would have a contradiction. Step 5: Follows since either \(a,b\) are both divisible by \(3\) (contradicting Step 2), or neither is, but then \(a^2\) and \(b^2\) are both one more than a multiple of \(3\) and the RHS is one more than a multiple of \(3\) but the LHS is \(2\) more than a multiple of \(3\) which is a contradiction.
  2. Step 1: If \(a\) is not divisible by \(5\) then \(a^2 \equiv \pm 1 \pmod{5}\) Step 2: Suppose \(\frac{a}{b} = \sqrt{6}+\sqrt{7}\) Step 3: \begin{align*} && \frac{a}{b} &= \sqrt{6}+\sqrt{7} \\ \Rightarrow && \frac{a^2}{b^2} &= 13 + 2\sqrt{42} \\ \Rightarrow && 168 &= \left (\frac{a^2}{b^2} - 13 \right)^2 \\ &&&= 169 - 26 \frac{a^2}{b^2} + \frac{a^4}{b^4} \\ \Rightarrow && a^4+b^4 &= 26a^2b^2 \end{align*} Step 4: If \(a\) is a multiple of \(5\) then so is \(b^4\) and hence so is \(b^2\) and \(b\). Step 5: But the left hand side is always \(2 \pmod{5}\) and the right hand side is never \(2 \pmod{5}\) contradiction. Divisibility by \(3\) doesn't work here since mod \(3\) we can have \(a = 1, b = 1\) and have a valid solution.

2018 Paper 2 Q6
D: 1600.0 B: 1484.7

  1. Find all pairs of positive integers \((n,p)\), where \(p\) is a prime number, that satisfy \[ n!+ 5 =p \,. \]
  2. In this part of the question you may use the following two theorems:
    1. For \(n\ge 7\), \(1! \times 3! \times \cdots \times (2n-1)! > (4n)!\,\).
    2. For every positive integer \(n\), there is a prime number between \(2n\) and \(4n\).
    Find all pairs of positive integers \((n,m)\) that satisfy \[ 1! \times 3! \times \cdots \times (2n-1)! = m! \,. \]


Solution:

  1. Let \(n! + 5 = p\). If \(n \geq 5\) then \(5\) divides the LHS and so must also divide the RHS. Since \(n!+5 > 5\) this means the RHS cannot be prime. Therefore consider \(n = 1, 2, 3, 4\). \begin{align*} n = 1: && 1! + 5 = 6 &&\text{ nope} \\ n=2: && 2! + 5 = 7 && \checkmark \\ n=3: && 3! + 5 = 11 && \checkmark \\ n=4: && 4! + 5 = 29 && \checkmark \end{align*} Therefore the solutions are \((2,7), (3,11), (4,29)\).
  2. Suppose \(1! \times 3! \times \cdots \times (2n-1)! = m!\). If \(n \geq 7\) then \(m! > (4n)!\) (by the first theorem) in particular \(m > 4n\). Therefore (by the second theorem) the RHS is divisible by some prime which cannot divide the LHS. Therefore consider \(n = 1,2,3,4,5,6\) \begin{align*} n = 1: && 1! = 1 = 1! && \checkmark \\ n = 2: && 1! \times 3! = 6 = 3! && \checkmark \\ n = 3: && 1! \times 3! \times 5! = 6! && \checkmark \\ n = 4: && 1! \times 3! \times 5! \times 7! = 6! \times 7! = 10! && \checkmark \\ n = 5: && 1! \times 3! \times 5! \times 7! \times 9! = 10! 9! > 11! && \text{would need a factor of } 11\text{ so no} \\ n = 6: && 1! \times 3! \times 5! \times 7! \times 9! \times 11! = 10! 11! 9! > 13! && \text{would need a factor of } 13\text{ so no} \\ \end{align*} Therefore all solutions are \((1,1), (2,3), (3,6), (4,10)\)

2016 Paper 1 Q7
D: 1500.0 B: 1500.0

The set \(S\) consists of all the positive integers that leave a remainder of 1 upon division by 4. The set \(T\) consists of all the positive integers that leave a remainder of 3 upon division by 4.

  1. Describe in words the sets \(S \cup T\) and \(S \cap T\).
  2. Prove that the product of any two integers in \(S\) is also in \(S\). Determine whether the product of any two integers in \(T\) is also in \(T\).
  3. Given an integer in \(T\) that is not a prime number, prove that at least one of its prime factors is in \(T\).
  4. For any set \(X\) of positive integers, an integer in \(X\) (other than 1) is said to be \(X\)-prime if it cannot be expressed as the product of two or more integers all in \(X\) (and all different from 1).
    • \(\bf (a)\) Show that every integer in \(T\) is either \(T\)-prime or is the product of an odd number of \(T\)-prime integers.
    • \(\bf (b)\) Find an example of an integer in \(S\) that can be expressed as the product of \(S\)-prime integers in two distinct ways. [Note: \(s_1s_2\) and \(s_2s_1\) are not counted as distinct ways of expressing the product of \(s_1\) and \(s_2\).]


Solution:

  1. \(S \cup T\) is the set of odd positive integers. \(S \cap T\) is the empty set.
  2. Suppose we have two integers in \(S\), say \(4n+1\) and \(4m+1\), so \((4n+1)(4m+1) = 16nm + 4(n+m) + 1 = 4(4nm + n+m) + 1\) which clearly is in \(S\). The product of any two integers in \(T\) is in \(S\), since \((4n+3)(4m+3) = 16nm + 12(n+m) + 9 = 4(4nm+3(n+m) + 2) + 1\)
  3. Suppose \(p \in T\) is not prime, then it must have prime factors. They must all be odd, so they are in \(T\) or \(S\). If they are all in \(S\) then \(p\) must also be in \(S\) (as the product of numbers in \(S\)) but this is a contraction. Therefore \(p\) must have a prime factor in \(T\).
    • \(3, 7, 11, 19\) are all primes in \(T\). Notice therefore that any product of two of them must be \(S\)-prime. But then \((3 \times 7) \times (11 \times 19)\) and \((3 \times 11) \times (7 \times 19)\) are distinct factorisations into \(S\)-prime factors.

2016 Paper 3 Q5
D: 1700.0 B: 1500.0

  1. By considering the binomial expansion of \((1+x)^{2m+1}\), prove that \[ \binom{ 2m \! +\! 1}{ m} < 2^{2m}\,, \] for any positive integer \(m\).
  2. For any positive integers \(r\) and \(s\) with \(r< s\), \(P_{r,s}\) is defined as follows: \(P_{r,s}\) is the product of all the prime numbers greater than \(r\) and less than or equal to \(s\), if there are any such primes numbers; if there are no such primes numbers, then \(P_{r,s}=1\,\). For example, \(P_{3,7}=35\), \(P_{7,10}=1\) and \(P_{14,18}=17\). Show that, for any positive integer \(m\), \(P_{m+1\,,\, 2m+1} \) divides \(\displaystyle \binom{ 2m \! +\! 1}{ m} \,,\) and deduce that \[ P_{m+1\,,\, 2m+1} < 2^{2m} \,. \]
  3. Show that, if \(P_{1,k} < 4^k\) for \(k = 2\), \(3\), \(\ldots\), \(2m\), then \( P_{1,2m+1} < 4^{2m+1}\,\).
  4. Prove that \(\P_{1,n} < 4^n\) for \(n\ge2\).


Solution:

  1. Notice that \((1+x)^{2m+1} = 1+\binom{2m+1}{1}x+\cdots + \binom{2m+1}{m}x^{m} + \binom{2m+1}{m+1} + \cdots\). Notice also that \(\binom{2m+1}{m} = \binom{2m+1}{m+1}\). Therefore evaluating at \(x = 1\), we see \(2^{2m+1} > \binom{2m+1}{m} + \binom{2m+1}{m+1} = 2 \binom{2m+1}{m} \Rightarrow \binom{2m+1}{m} < 2^{2m}\)
  2. Each prime dividing \(P_{m+1, 2m+1}\) divides the numerator of \(\binom{2m+1}{m}\) since it appears in \((2m+1)!\), but not the denominator, since they wont appear in \(m!\) or \((m+1)!\), and since they are prime they have to appear to divide it. Therefore the must divide \(\binom{2m+1}{m}\) and therefore \(P_{m+1,2m+1}\) must divide that binomail coefficient. Since \(a \mid b \Rightarrow a \leq b\) we must have \(P_{m+1, 2m+1} \leq \binom{2m+1}{m} < 2^{2m}\)
  3. Since \begin{align*} P_{1,2m+1} &= P_{1,m+1}P_{m+1, 2m+1} \tag{split into primes below \(m+1\) and abvoe} \\ &< 4^{m+1}P_{m+1,2m+1} \tag{use the condition from the question}\\ &<4^{m+1}2^{2m} \tag{use our inequality} \\ &= 4^{2m+1} \end{align*}
  4. We proceed by (strong) induction. Base case: (\(n = 2\)): \(P_{1,2} = 2 < 4^2 =16\) Suppose it is true for all \(k=2,3,\cdots,2m\) then it is true for \(k=2m+1\) by the previous part of the question. However it is also true for \(k=2m+2\), since that can never be prime (as n is now an even number bigger than 2). Therefore by the principle of mathematical induction it is true for all \(n\).

2014 Paper 1 Q1
D: 1500.0 B: 1500.0

All numbers referred to in this question are non-negative integers.

  1. Express each of the numbers 3, 5, 8, 12 and 16 as the difference of two non-zero squares.
  2. Prove that any odd number can be written as the difference of two squares.
  3. Prove that all numbers of the form \(4k\), where \(k\) is a non-negative integer, can be written as the difference of two squares.
  4. Prove that no number of the form \(4k+2\), where \(k\) is a non-negative integer, can be written as the difference of two squares.
  5. Prove that any number of the form \(pq\), where \(p\) and \(q\) are prime numbers greater than 2, can be written as the difference of two squares in exactly two distinct ways. Does this result hold if \(p\) is a prime greater than 2 and \(q=2\)?
  6. Determine the number of distinct ways in which 675 can be written as the difference of two squares.


Solution:

  1. \(\,\) \begin{align*} && 3 &= 2^2 - 1^2 \\ && 5 &= 3^2 - 2^2 \\ && 8 &= 3^2 - 1^2 \\ && 16 &= 5^2 - 3^2 \end{align*}
  2. Suppose \(n = 2k+1\), then \(n = (k+1)^2 - k^2\)
  3. Suppose \(n = 4k\) then \(n = (2k+1)^2 - (2k-1)^2\)
  4. All squares leave a remainder of \(0\) or \(1\) on division by \(4\). Therefore the difference can leave a remainder of \(0\), \(1\), \(-1 \equiv 3\), none of which are \(2\).
  5. Suppose \(n = pq = a^2 - b^2\) with \(a > b\) ie \((a-b)(a+b) = pq\). Since \(p\) is prime, \(p \mid (a-b)\) or \(p \mid (a+b)\). Similarly for \(q\). Suppose also (wlog) that \(p > q\) Since the factors of \(pq\) are \(1, p, q, pq\) then \(a-b = 1, p\) (which are two possibilities) and \(a+b = pq, q\), ie \(a = \frac{1+pq}{2}, \frac{p+q}{2}\) and \(b = \frac{pq-1}{2}, \frac{p-q}{2}\) \begin{align*} && pq &= \left ( \frac{1+pq}{2} \right)^2- \left ( \frac{1-pq}{2} \right)^2 \\ &&&= \left ( \frac{p+q}{2} \right)^2- \left ( \frac{p-q}{2} \right)^2 \\ \end{align*} Where everything is an integer since \(p\) and \(q\) are odd. If we have \(p > 2\) and \(q = 2\) then \(p\) is odd and the number has the form \(4k+2\) which cannot be expressed as the difference of two squares.
  6. \(675 = 3^3 \cdot 5^2\), each factor pair of \(675\) will lead to a different solution of \(675 = a^2-b^2\), since we will have an equation \(a-b = X, a+b = Y\) where \(X, Y\) are both odd. Therefore there are as many solution as (half) the number of factors, ie \(4 \times 3 = 12\)

2013 Paper 2 Q7
D: 1600.0 B: 1516.0

  1. Write down a solution of the equation \[ x^2-2y^2 =1\,, \tag{\(*\)} \] for which \(x\) and \(y\) are non-negative integers. Show that, if \(x=p\), \(y=q\) is a solution of (\(*\)), then so also is \(x=3p+4q\), \(y=2p+3q\). Hence find two solutions of \((*)\) for which \(x\) is a positive odd integer and \(y\) is a positive even integer.
  2. Show that, if \(x\) is an odd integer and \(y\) is an even integer, \((*)\) can be written in the form \[ n^2 = \tfrac12 m(m+1)\,, \] where \(m\) and \(n\) are integers.
  3. The positive integers \(a\), \(b\) and \(c\) satisfy \[ b^3=c^4-a^2\,, \] where \(b\) is a prime number. Express \(a\) and \(c^2\) in terms of \(b\) in the two cases that arise. Find a solution of \(a^2+b^3=c^4\), where \(a\), \(b\) and \(c\) are positive integers but \(b\) is not prime.


Solution:

  1. \((x,y) = (1,0)\) we have Suppose \(p^2-2q^2 = 1\), then \begin{align*} && (3p+4q)^2-2\cdot(2p+3q)^2 &= 9p^2+24pq + 16q^2 - 2\cdot(4p^2+12pq+9q^2) \\ &&&= p^2(9-8) + pq(24-24) + q^2(16-18) \\ &&&= p^2 - 2q^2 = 1 \end{align*} So we have: \begin{array}{c|c} x & y \\ \hline 1 & 0 \\ 3 & 2 \\ 17 & 12 \\ \end{array}
  2. Suppose \(x = 2m+1\) and \(y = 2n\) then \begin{align*} && 1 & = x^2 - 2y^2 \\ &&&= (2m+1)^2 - 2(2n)^2 \\ &&&= 4m^2 + 4m + 1 - 8n^2 \\ \Leftrightarrow && n^2 &= \frac{m(m+1)}{2} \end{align*}
  3. Suppose \(b^3 = c^4 - a^2 =(c^2-a)(c^2+a)\), since \(b\) is prime and \(c^2 + a > c^2-a\) we must have: \begin{align*} && p = c^2-a && p^2 =c^2 +a \\ \Rightarrow && c^2 = \frac{p+p^2}{2} && a = \frac{p^2-p}2\\ && 1 = c^2-a && p^3 = c^2+a \\ \Rightarrow && c^2 = \frac{p^3+1}{2} && a = \frac{p^3-1}{2} \end{align*} Note that \(c^2 = \frac{p(p+1)}{2}\) is reminicent of our first equation, so suppose \(n = c = 6\) and \(p = m = 8\) then \(6^4 = 8^3 + 28^2\)

2013 Paper 3 Q5
D: 1700.0 B: 1487.0

In this question, you may assume that, if \(a\), \(b\) and \(c\) are positive integers such that \(a\) and \(b\) are coprime and \(a\) divides \(bc\), then \(a\) divides \(c\). (Two positive integers are said to be coprime if their highest common factor is 1.)

  1. Suppose that there are positive integers \(p\), \(q\), \(n\) and \(N\) such that \(p\) and \(q\) are coprime and \(q^nN=p^n\). Show that \(N=kp^n\) for some positive integer \(k\) and deduce the value of \(q\). Hence prove that, for any positive integers \(n\) and \(N\), \(\sqrt[n]N\) is either a positive integer or irrational.
  2. Suppose that there are positive integers \(a\), \(b\), \(c\) and \(d\) such that \(a\) and \(b\) are coprime and \(c\) and \(d\) are coprime, and \(a^ad^b = b^a c^b \,\). Prove that \(d^b = b^a\) and deduce that, if \(p\) is a prime factor of \(d\), then \(p\) is also a prime factor of \(b\). If \(p^m\) and \(p^n\) are the highest powers of the prime number \(p\) that divide \(d\) and \(b\), respectively, express \(b\) in terms of \(a\), \(m\) and \(n\) and hence show that \(p^n\le n\). Deduce the value of \(b\). (You may assume that if \(x > 0\) and \(y\ge2\) then \(y^x > x\).) Hence prove that, if \(r\) is a positive rational number such that \(r^r\) is rational, then \(r\) is a positive integer.


Solution:

  1. Suppose \(q^nN = p^n\) then since \((p,q) =1\) we must have \(p \mid N\), and then by dividing both \(p^n\) and \(N\) by \(p\) we can repeat this process \(n\) times to find that \(N = kp^n\) and in particular \(q = 1\). Suppose \(\sqrt[n]{N} = \frac{p}q\) for \(p,q\) coprime positive integers (ie it is not irrational), then \(q^nN = p^n\) and so \(q = 1\) and in fact \(\sqrt[n]{N}\) is an integer so \(N\).
  2. Suppose \((a,b) = 1, (c,d) = 1\) and \(a^ad^b = b^ac^b\), then since \((a,b) = 1\) we must have \((b^a, a) = 1\) so \(b^a \mid d^b\). Similarly since \((c,d) = 1\) we must have \((d^b, c) = 1\) so \(d^b \mid b^a\). Therefore \(d^b = b^a\). Suppose \(p \mid d\) then \(p \mid d^b = b^a \Rightarrow p \mid b\). Suppose \(\nu_p(d) = m, \nu_p(b) = n\) we must have \(bm = \nu_p(d^b) = \nu_p(b^a) = an\), ie \(b = \frac{an}{m}\). Note that \(p^n \mid b \Rightarrow p^n \mid n \frac{a}{m} \Rightarrow p^n \mid n \Rightarrow p^n \leq n\). Since \((p,a) = 1\).. But since \(p^n > n\) if \(p \geq 2\) we must have that \(b = 1\). Therefore suppose \(r = \frac{a}{b}\) with \((a,b) = 1\) an \(r^r = \frac{c}{d}\) we must ahve \(a^ac^b = b^ad^b\) and so \(b = 1\) implying \(r\) is an integer.

2012 Paper 3 Q5
D: 1700.0 B: 1554.6

  1. The point with coordinates \((a, b)\), where \(a\) and \(b\) are rational numbers,is called an integer rational point if both \(a\) and \(b\) are integers; a non-integer rational point if neither \(a\) nor \(b\) is an integer.
    • \(\bf (a)\) Write down an integer rational point and a non-integer rational point on the circle \(x^2+y^2 =1\).
    • [\bf (b)] Write down an integer rational point on the circle \(x^2+y^2=2\). Simplify \[ (\cos\theta + \sqrt m \sin\theta)^2 + (\sin\theta - \sqrt m \cos\theta)^2 \, \] and hence obtain a non-integer rational point on the circle \(x^2+y^2=2\,\).
  2. The point with coordinates \((p+\sqrt 2 \, q\,,\, r+\sqrt 2 \, s)\), where \(p\), \(q\), \(r\) and \(s\) are rational numbers, is called: an integer \(2\)-rational point if all of \(p\), \(q\), \(r\) and \(s\) are integers; a non-integer \(2\)-rational point if none of \(p\), \(q\), \(r\) and \(s\) is an integer.
    • \(\bf (a)\) Write down an integer \(2\)-rational point, and obtain a non-integer \(2\)-rational point, on the circle \(x^2+y^2=3\,\).
    • [\bf(b)] Obtain a non-integer \(2\)-rational point on the circle \(x^2+y^2=11\,\).
    • [\bf(c)]Obtain a non-integer \(2\)-rational point on the hyperbola \(x^2-y^2 =7 \).


Solution:

    • \(\bf (a)\) \((1, \sqrt2)\) is an integer \(2\)-rational point. \((\frac35 + \frac45\sqrt2, \frac45 - \frac{3}{5}\sqrt2)\) is a non-integer \(2\)-rational point.
    • [\bf(b)] First notice that \((\sqrt2)^2 +3^2 = 11\) so then consider \((1 + \tfrac32\sqrt2, 1-\tfrac32\sqrt2)\) will work as \(\pi/4\) degree rotation.
    • [\bf(c)] First notice \(3^2-(\sqrt2)^2 = 2\). Notice that \((k\sec \theta + \sqrt{m} \tan \theta)^2 - (k\tan \theta + \sqrt{m} \sec \theta)^2 = k^2-m\). Taking \(k= 3\) we have \((3 \cdot \frac{13}{5} + \frac{12}{5}\sqrt{2}, 3\cdot\frac{12}5+\frac{13}{5}\sqrt2)\)
Note: we can also find the additional point in the last part by considering lines through \((3, \sqrt2)\), for example \(y = -\frac32x + \sqrt2 + \frac92\) would give the same point.