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2022 Paper 3 Q2
D: 1500.0 B: 1500.0
number theory proof by contradiction infinite descent divisibility diophantine equation modular arithmetic

  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 \(a^2 + b^2 + c^2 = 4abc\).

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2010 Paper 1 Q8
D: 1500.0 B: 1484.0
proof by contradiction divisibility infinite descent number theory modular arithmetic integer solutions proof

  1. Suppose that \(a\), \(b\) and \(c\) are integers that satisfy the equation \[ a^{3}+3b^{3}=9c^{3}. \] Explain why \(a\) must be divisible by 3, and show further that both \(b\) and \(c\) must also be divisible by 3. Hence show that the only integer solution is \(a=b=c=0\,\).
  2. Suppose that \(p\), \(q\) and \(r\) are integers that satisfy the equation \[ p^4 +2q^4 = 5r^4 \,.\] By considering the possible final digit of each term, or otherwise, show that \(p\) and \(q\) are divisible by 5. Hence show that the only integer solution is \(p=q=r=0\,\).

Solution:

  1. Since \(a^3 = 9c^3 - 3b^3 = 3(3c^3-b^3)\) we must have \(3 \mid a^3\). But since \(3\) is prime, \(3 \mid a\). Since \(3 \mid a\) we can write \(a = 3a'\) for some \(a' \in \mathbb{Z}\). Therefore our equation is \(27(a')^3 + 3b^3 = 9c^3 \Rightarrow 9(a')^3 + b^3 = 3c^3\) which means that \(3 \mid b\) by the same argument from earlier. So \(b = 3b'\) so the equation is \(9(a')^3 + 27(b')^3 = 3c^3 \Rightarrow 3(a')^3 + 9(b')^3 = c^3\) which means that \(3 \mid c\). Suppose \((a,b,c)\) is the smallest measured by \(a^2+b^2+c^2\) with \(a, b, c\neq 0\). Then \((\frac{a}{3}, \frac{b}{3}, \frac{c}{3})\) is also a solution. But this contradicts that we had found the smallest solution. Therefore the only possible solution is \((0,0,0)\) which clearly works.
  2. Consider \(p, q \pmod{5}\). By \(FLT\) \(p^4, q^4 = 0, 1 \pmod{5}\) so \(p^4+2q^4 \in \{0, 1, 2, 3\}\) and in particular the only way they are divisible by \(5\) is if \(p \equiv q \equiv 0 \pmod{5}\). Therefore \(p = 5p', q = 5q'\) and so \(5^4(p')^4 + 5^4(q')^4 = 5r^4 \Rightarrow r^4 = 5(25(p')^4 + 25(q')^4) \Rightarrow 5\mid r^4 \Rightarrow 5 \mid r\). Therefore we can use the same argument about the smallest solution to show that \(p = q= r = 0\)

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1993 Paper 2 Q7
D: 1600.0 B: 1491.2
number theory modular arithmetic divisibility proof by contradiction infinite descent Diophantine equation mod 5

The integers \(a,b\) and \(c\) satisfy \[ 2a^{2}+b^{2}=5c^{2}. \] By considering the possible values of \(a\pmod5\) and \(b\pmod5\), show that \(a\) and \(b\) must both be divisible by \(5\). By considering how many times \(a,b\) and \(c\) can be divided by \(5\), show that the only solution is \(a=b=c=0.\)

Solution: \begin{array}{c|ccccc} a & 0 & 1 & 2 & 3 & 4 \\ a^2 & 0 & 1 & 4 & 4 & 1 \end{array} Therefore \(a^2 \in \{0,1,4\}\) and so we can have \begin{array} $2a^2+b^2 & 0 & 1 & 4 \\ \hline 0 & 0 & 1 & 4 \\ 1 & 2 & 3 & 1 \\ 4 & 3 & 4 & 2 \end{array} Therefore the only solution must have \(5 \mid a,b\), but then we can write them has \(5a'\) and \(5b'\) so the equation becomes \(2\cdot25 a'^2 + 25b'^2 = 5c^2\) ie \(5 \mid c^2 \Rightarrow 5 \mid c\). But that means we can always divide \((a,b,c)\) by \(5\), which is clearly a contradiction if we consider the lowest power of \(5\) dividing \(a,b,c\) for any solution.

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