Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 4

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 $u_n\in \mathbb{Z}$.

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 $2^n$.

The congruences reduce to $z\equiv 14(\mathrm{mod}17)$ and $z\equiv 8(\mathrm{mod}19)$. Writing $z=14+17k$ gives $k\equiv 3(\mathrm{mod}19)$, hence $z\equiv 65(\mathrm{mod}323)$.

$\phi (33)=20$ and $7^{-1}\equiv 3(\mathrm{mod}20)$, so Bob computes $29^3\equiv 2(\mathrm{mod}33)$. RSA relies on modular exponentiation being easy while recovering the private exponent without the factorisation of a large $N$ is believed hard.

From $F=G{m}_1{m}_2/r^2$, $[G]=L^3{M}^{-1}{T}^{-2}$.

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

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

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

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

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

Whether $\log 2$ is irrational is not presently known; the given fact about $e$ does not decide it.

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

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

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 $\pm 1,\pm 3$ gives none, so the root is irrational.

Let $2^k\le n<2^{k+1}$. 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.

For $a\ne 0(\mathrm{mod}p)$, $\gcd (a,p)=1$, so Bézout gives $au+pv=1$ and $u$ is an inverse modulo $p$.

If $x^2\equiv a$ and $y^2\equiv a(\mathrm{mod}p)$, then $(x-y)(x+y)\equiv 0$, hence $y\equiv \pm x$; these differ for odd $p$. It fails for composite moduli: $1$ has four square roots modulo $8$.

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

Wilson gives $18!\equiv -1(\mathrm{mod}19)$. Since $18\cdot 17\equiv 2$, $2\cdot 16!\equiv -1$, hence $16!\equiv 9(\mathrm{mod}19)$.

By Fermat, $a^{p-1}=1$. Thus $a^{(p-1)/2}=\pm 1$. The squaring map on ${\mathbb{F}}_p^{\ast }$ has kernel $\{\pm 1\}$, 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.

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

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

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 $f(A)$.

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

Put $|B|=m$. There are $\left(\genfrac{}{}{0ex}{}{m+1}{2}\right)m!=(m+1)!m/2$ surjections and $(m+1)!$ injections. For $m\ge 3$ 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.

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.

$X_1$ is the set of real algebraic numbers and is countable because each nonzero polynomial has finitely many roots. For $n\ge 2$, $X_n$ is uncountable: the polynomial $x_1$ vanishes on $\{0\}\times {\mathbb{R}}^{n-1}$.

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 $S={\bigcup }_n{S}_n$ is countable and has (i)–(iii). Any other set with those properties contains every $S_n$ by induction, proving minimality and uniqueness.

Let the full cube mass be $M=8\rho a^3$ and the removed cylinder mass $m_h=2\pi \rho a{r}^2$. About the rod, the remaining moment isThe remaining centre-of-mass distance $\ell$ obeys $(M-m_h)\ell =M\gamma a$, so the small-angle equation is $I\ddot{\theta }+Mg\gamma a\theta =0$. Hence

$m\ddot{x}=-\nabla V$. Dotting with $\dot{x}$ gives $d(\frac{1}{2}m|\dot{x}{|}^2+V)/dt=0$.

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 $L$.

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 $\cos \alpha =-q\beta /|J|$.

$L^2=J^2-(q\beta {)}^2=J^2{\sin }^2\alpha$. Splitting kinetic energy into radial and angular parts therefore gives $E=\frac{1}{2}m{\dot{r}}^2+J^2{\sin }^2\alpha /(2m{r}^2)+V(r)$, proving the stated effective potential.

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

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

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 $v_e=c(m_{\pi }^2-m_e^2)/(m_{\pi }^2+m_e^2)$.

Let the incident positron total energy be $E$. For perpendicular photon momenta, real positive photon energies require
$(E+m_e{c}^2{)}^2-4m_e{c}^2(E+m_e{c}^2)\ge 0$.
Thus the threshold is $E=3m_e{c}^2$.

Squaring $p_H=p_1+p_2$ gives $m^2{c}^4=2E_1{E}_2(1-\cos \theta )=4E_1{E}_2{\sin }^2(\theta /2)$, 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 $v=c\cos (\theta /2)$.