past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-4.bigb

= Paper 4
{scope}

https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2026/Paperia_4_2026.pdf

= 1F
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= i
{parent=1f}
{scope}

= Solution
{parent=i}

The two conjugate roots satisfy $r^2=6r-4$, so $u_n=6u_{n-1}-4u_{n-2}$ with $u_0=2,u_1=6$. Induction gives $u_n\in\mathbb Z$.

Solved by gpt-5.6-sol high.

= ii
{parent=1f}
{scope}

= Solution
{parent=ii}

The same recurrence and $2^0\mid u_0$, $2^1\mid u_1$ prove the claim: if $2^{n-1}\mid u_{n-1}$ and $2^{n-2}\mid u_{n-2}$, both terms on the right are divisible by $2^n$.

Solved by gpt-5.6-sol high.

= 2D
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= a
{parent=2d}
{scope}

= Solution
{parent=a}

The congruences reduce to $z\equiv14\pmod {17}$ and $z\equiv8\pmod {19}$. Writing $z=14+17k$ gives $k\equiv3\pmod {19}$, hence $z\equiv65\pmod {323}$.

Solved by gpt-5.6-sol high.

= b
{parent=2d}
{scope}

= Solution
{parent=b}

$\varphi(33)=20$ and $7^{-1}\equiv3\pmod {20}$, so Bob computes $29^3\equiv2\pmod {33}$. RSA relies on modular exponentiation being easy while recovering the private exponent without the factorisation of a large $N$ is believed hard.

Solved by gpt-5.6-sol high.

= 3B
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= i
{parent=3b}
{scope}

= Solution
{parent=i}

From $F=Gm_1m_2/r^2$, $[G]=L^3M^{-1}T^{-2}$.

Solved by gpt-5.6-sol high.

= ii
{parent=3b}
{scope}

= Solution
{parent=ii}

Using only $G,M$ and orbital scale $R$, dimensional balance gives $T^2\propto R^3/(GM)$.

Solved by gpt-5.6-sol high.

= iii
{parent=3b}
{scope}

= Solution
{parent=iii}

The only length from $G,M,c$ is $R\sim GM/c^2$.

Solved by gpt-5.6-sol high.

= iv
{parent=3b}
{scope}

= Solution
{parent=iv}

The only length from $\hbar,m,c$ is $\lambda\sim\hbar/(mc)$.

Solved by gpt-5.6-sol high.

= v
{parent=3b}
{scope}

= Solution
{parent=v}

Equating the two lengths gives $m\sim\sqrt{\hbar c/G}$, the <Planck mass>.

Solved by gpt-5.6-sol high.

= 4B
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= Solution
{parent=4b}

<Proper time> satisfies $c^2d\tau^2=c^2dt^2-dx^2$ and is invariant, so differentiating by it defines a four-vector. The path is timelike iff $R|\omega|\lt c$. Its next return occurs after coordinate time $2\pi/|\omega|$, so Bob ages
$$\Delta\tau=\frac{2\pi}{|\omega|}\sqrt{1-R^2\omega^2/c^2}.$$
With $\gamma=(1-R^2\omega^2/c^2)^{-1/2}$,
$U^\mu=\gamma(c,R\omega\cos\omega t,-R\omega\sin\omega t,0)$ and $U\cdot U=c^2$ for signature $(+---)$.

Solved by gpt-5.6-sol high.

= 5F
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= i
{parent=5f}
{scope}

= a
{parent=i}
{scope}

= Solution
{parent=a}

Whether $\log2$ is irrational is not presently known; the given fact about $e$ does not decide it.

Solved by gpt-5.6-sol high.

= b
{parent=i}
{scope}

= Solution
{parent=b}

$\log_2 3$ is irrational: if it were $p/q$, then $2^p=3^q$, contradicting unique prime factorisation.

Solved by gpt-5.6-sol high.

= c
{parent=i}
{scope}

= Solution
{parent=c}

$ae+b/e$ is transcendental, hence irrational. If it were algebraic, $e$ would solve $ax^2-(ae+b/e)x+b=0$ over the algebraic numbers, contradicting transcendence of $e$.

Solved by gpt-5.6-sol high.

= d
{parent=i}
{scope}

= Solution
{parent=d}

The cubic is strictly increasing and has one real root. A rational root of the monic integer <polynomial> would be an integer divisor of $3$; testing $\pm1,\pm3$ gives none, so the root is irrational.

Solved by gpt-5.6-sol high.

= ii
{parent=5f}
{scope}

= Solution
{parent=ii}

Let $2^k\le n\lt 2^{k+1}$. In the reduced common-denominator expression for $H_n$, the term $1/2^k$ is the unique term whose denominator contains the largest power $2^k$; after multiplication by the least common multiple it contributes an odd integer while all other terms contribute even integers. Thus the numerator is odd and the denominator remains even, so $H_n$ is not an integer.

Solved by gpt-5.6-sol high.

= 6D
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= i
{parent=6d}
{scope}

= Solution
{parent=i}

For $a\ne0\pmod p$, $\gcd(a,p)=1$, so Bézout gives $au+pv=1$ and $u$ is an inverse modulo $p$.

Solved by gpt-5.6-sol high.

= ii
{parent=6d}
{scope}

= Solution
{parent=ii}

If $x^2\equiv a$ and $y^2\equiv a\pmod p$, then $(x-y)(x+y)\equiv0$, hence $y\equiv\pm x$; these differ for odd $p$. It fails for composite <moduli>: $1$ has four square roots modulo $8$.

Solved by gpt-5.6-sol high.

= iii
{parent=6d}
{scope}

= Solution
{parent=iii}

For prime $n$, pair each nonzero residue with its inverse; only $\pm1$ are self-inverse, yielding <Wilson theorem>. Conversely, if $(n-1)!\equiv-1\pmod n$, every $1\le a\lt n$ is a unit (otherwise a common divisor would divide the left side and $n$ but not $-1$), so $n$ is prime.

Solved by gpt-5.6-sol high.

= iv
{parent=6d}
{scope}

= Solution
{parent=iv}

Wilson gives $18!\equiv-1\pmod {19}$. Since $18\cdot17\equiv2$, $2\cdot16!\equiv-1$, hence $16!\equiv9\pmod {19}$.

Solved by gpt-5.6-sol high.

= v
{parent=6d}
{scope}

= Solution
{parent=v}

By Fermat, $a^{p-1}=1$. Thus $a^{(p-1)/2}=\pm1$. The squaring map on $\mathbb F_p^*$ has kernel $\{\pm1\}$, so exactly $(p-1)/2$ inputs are squares; the <polynomial> $x^{(p-1)/2}-1$ has at most that many roots and already has all squares, proving the two cases.

Solved by gpt-5.6-sol high.

= 7E
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= a
{parent=7e}
{scope}

= Solution
{parent=a}

True: take $C=f(A)$ and define $g(f(a))=a$, which is well-defined and injective because $f$ is injective.

Solved by gpt-5.6-sol high.

= b
{parent=7e}
{scope}

= Solution
{parent=b}

False: a constant surjection from a two-element set to a singleton followed by the singleton’s inclusion is not injective.

Solved by gpt-5.6-sol high.

= c
{parent=7e}
{scope}

= Solution
{parent=c}

False: include a singleton into a two-element set and then map both elements onto a singleton; the composite need not cover a larger target in analogous examples. Concretely $A=\{1\}$, $B=\{1,2\}$, $C=\{1,2\}$, $f(1)=1$, and let $g$ swap/surject with $g(1)=g(2)=1$ after taking $C$ singleton; then the claimed implication fails whenever an element of $C$ is reached only outside $f(A)$.

Solved by gpt-5.6-sol high.

= d
{parent=7e}
{scope}

= Solution
{parent=d}

True: if $f(a_1)=f(a_2)$ with $a_1\ne a_2$, then $(g\circ f)(a_1)=(g\circ f)(a_2)$.

Solved by gpt-5.6-sol high.

= e
{parent=7e}
{scope}

= Solution
{parent=e}

Put $|B|=m$. There are $\binom{m+1}{2}m!=(m+1)!m/2$ surjections and $(m+1)!$ injections. For $m\ge3$ the domain is larger, so $\Psi$ cannot be injective, regardless of the choices. For $m=1$ there is only one surjection, so it is injective. For $m=2$ both sets have size $6$; injectivity depends on the representative choices and, if achieved, is equivalent to bijectivity.

Solved by gpt-5.6-sol high.

= 8E
{parent=Paper 4}
{scope}
{title2=Numbers and Sets}

= a
{parent=8e}
{scope}

= Solution
{parent=a}

Monomials are indexed by $\mathbb N^n$, a <countable set>. A <polynomial> is a finite list of monomials with rational coefficients, so $P_n$ is a countable union of countable sets.

Solved by gpt-5.6-sol high.

= b
{parent=8e}
{scope}

= Solution
{parent=b}

$X_1$ is the set of real algebraic numbers and is countable because each nonzero <polynomial> has finitely many roots. For $n\ge2$, $X_n$ is uncountable: the <polynomial> $x_1$ vanishes on $\{0\}\times\mathbb R^{n-1}$.

Solved by gpt-5.6-sol high.

= c
{parent=8e}
{scope}

= Solution
{parent=c}

Start with $S_0=\mathbb Q$. Given countable $S_n$, let $S_{n+1}$ be the field generated by $S_n\cup\{\sqrt{|x|}:x\in S_n\}$. It is countable because its elements are values of rational expressions in finitely many members of a countable set. Then $S=\bigcup_nS_n$ is countable and has (i)–(iii). Any other set with those properties contains every $S_n$ by induction, proving minimality and uniqueness.

Solved by gpt-5.6-sol high.

= 9B
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= Solution
{parent=9b}

Let the full cube mass be $M=8\rho a^3$ and the removed cylinder mass $m_h=2\pi\rho ar^2$. About the rod, the remaining moment is
$$I=M a^2(2/3+\gamma^2)-\tfrac12m_hr^2.$$
The remaining centre-of-mass distance $\ell$ obeys $(M-m_h)\ell=M\gamma a$, so the small-angle equation is $I\ddot\theta+M g\gamma a\,\theta=0$. Hence
$$T=2\pi\sqrt{\frac a{g\gamma}\left(\frac23+\gamma^2-\frac\pi8\frac{r^4}{a^4}\right)}.$$

Solved by gpt-5.6-sol high.

= 10B
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= a
{parent=10b}
{scope}

= Solution
{parent=a}

$m\ddot x=-\nabla V$. Dotting with $\dot x$ gives $d(\tfrac12m|\dot x|^2+V)/dt=0$.

Solved by gpt-5.6-sol high.

= b
{parent=10b}
{scope}

= Solution
{parent=b}

A central potential is $V(r)$. Then $\dot L=x\times(-\nabla V)=0$. Since $L\cdot x=0$, the trajectory lies in the fixed plane perpendicular to $L$.

Solved by gpt-5.6-sol high.

= c
{parent=10b}
{scope}

= Solution
{parent=c}

The equation is $m\ddot x=-\nabla V+q\dot x\times B$; the magnetic force does no work. For $B=\beta\hat r/r^2$, the triple-product identity gives $\dot L=q\beta\,d\hat r/dt$, so $J=L-q\beta\hat r$ is conserved. Since $J\cdot\hat r=-q\beta$, the trajectory lies on a cone with $\cos\alpha=-q\beta/|J|$.

Solved by gpt-5.6-sol high.

= d
{parent=10b}
{scope}

= Solution
{parent=d}

$L^2=J^2-(q\beta)^2=J^2\sin^2\alpha$. Splitting <kinetic energy> into radial and angular parts therefore gives $E=\tfrac12m\dot r^2+J^2\sin^2\alpha/(2mr^2)+V(r)$, proving the stated <effective potential>.

Solved by gpt-5.6-sol high.

= 11B
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= a
{parent=11b}
{scope}

= Solution
{parent=a}

Differentiating a <vector> using $(dA/dt)_S=(dA/dt)_{S\prime}+\omega\times A$ twice yields
$m\ddot x=F-2m\omega\times\dot x-m\dot\omega\times x-m\omega\times(\omega\times x)$.

Solved by gpt-5.6-sol high.

= b
{parent=11b}
{scope}

= Solution
{parent=b}

Along $BC$, write $x=(2a,s,0)$. With $\omega=-\alpha\hat z/t$, the tangential equation is
$$\ddot s=\frac{\alpha^2s-2a\alpha}{t^2}.$$
A stationary point is $s_0=2a/\alpha$, lying on the side iff $\alpha\ge1$. The general perturbed motion is
$$s=s_0+C t^{(1+\sqrt{1+4\alpha^2})/2}+D t^{(1-\sqrt{1+4\alpha^2})/2}.$$
Except on the decaying-mode fine tuning, the growing term carries the bead to $B$ or $C$, according to its sign.

Solved by gpt-5.6-sol high.

= 12B
{parent=Paper 4}
{scope}
{title2=Dynamics and Relativity}

= a
{parent=12b}
{scope}

= Solution
{parent=a}

Two-body energy-momentum conservation in the pion rest frame gives $E_e=(m_\pi^2+m_e^2)c^2/(2m_\pi)$ and $p_e=(m_\pi^2-m_e^2)c/(2m_\pi)$. Thus $v_e=c(m_\pi^2-m_e^2)/(m_\pi^2+m_e^2)$.

Solved by gpt-5.6-sol high.

= b
{parent=12b}
{scope}

= Solution
{parent=b}

Let the incident positron total energy be $E$. For perpendicular photon <momenta>, real positive photon energies require
$(E+m_ec^2)^2-4m_ec^2(E+m_ec^2)\ge0$.
Thus the threshold is $E=3m_ec^2$.

Solved by gpt-5.6-sol high.

= c
{parent=12b}
{scope}

= Solution
{parent=c}

Squaring $p_H=p_1+p_2$ gives $m^2c^4=2E_1E_2(1-\cos\theta)=4E_1E_2\sin^2(\theta/2)$, proving the formula. If the rest-frame photons are along $\pm y$, an $x$-boost gives equal energies and $\sin(\theta/2)=1/\gamma$, hence $v=c\cos(\theta/2)$.

Solved by gpt-5.6-sol high.