Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 4

The series $\sum a_n$ converges when its partial sums tend to a finite real limit. Put $m=n+1$. Thenby strict concavity of the square root. Taylor expansion at $m$ givesso $m^2{a}_n\to 1/8$. Limit comparison with $\sum n^{-2}$ proves convergence.

Every integer $n>1$ is a product of primes, uniquely up to order. Existence follows by strong induction: if $n$ is not prime, factor it into smaller positive integers and apply induction. For uniqueness, ifEuclid's lemma makes $p_1$ divide some $q_j$, hence equal it. Cancel that prime and induct on the number of factors.

Prime exponents must obey simultaneous congruences modulo $2,3,5$. The choicessatisfy the required congruences. Thus one example is

Newton's equation and its energy integral areThusLet $a=-V^{\prime \prime }(x_{+}^{\ast })>0$. Near the maximum,and the nearby turning point has distance $ϵ\sim \sqrt{2E^{\ast }\delta /a}$ from $x_{+}^{\ast }$. The singular part of the period is an $\mathrm{arcosh}$ integral:HencePhysically, the limiting orbit approaches the unstable equilibrium at the barrier top with vanishing speed and spends arbitrarily long there.

Put $R=R_A+R_B$, $c=\sqrt{R^2-b^2}/R$, and $s=b/R$. At impact the unit vector from A to B is $n=(-c,s)$. The impulse is normal, so tangential velocity is unchanged, while the normal components undergo a one-dimensional elastic collision. Therefore

If $\sqrt{2}+\sqrt{3}$ were rational, squaring would make $\sqrt{6}$ rational, a contradiction. Alsoso it is irrational but algebraic, satisfying $x^2+2x-1=0$.

For lattice vertices $u,v,w$, the triangle area isa half-integer. Triangulating a convex lattice polygon from one vertex makes its area rational.

A regular octagon of side length $s$ has area $2(1+\sqrt{2})s^2$. Lattice endpoints make $s^2$ a positive integer, so this area would be irrational, contradicting the preceding result. Hence no such octagon exists.

For pairwise coprime positive integers $n_i$, every system $x\equiv a_i(\mathrm{mod}{n}_i)$ has a unique solution modulo ${\prod }_i{n}_i$.

Euler's totient is $\varphi (n)=|\{1\le a\le n:(a,n)=1\}|$, equivalently the number of units modulo $n$.

If $(a,n)=1$, then $a^{\varphi (n)}\equiv 1(\mathrm{mod}n)$.

The Chinese remainder isomorphism $\mathbb{Z}/(nm)\cong \mathbb{Z}/n\times \mathbb{Z}/m$ restricts to a bijection on units. Counting those units gives $\varphi (nm)=\varphi (n)\varphi (m)$.

Write squarefree $n=\prod p$. For each $p\mid n$, either $p\mid a$, when both sides vanish modulo $p$, or Fermat's theorem applies. Since $p-1\mid \varphi (n)$ and $k\equiv 1(\mathrm{mod}\varphi (n))$, it gives $a^k\equiv a(\mathrm{mod}p)$. The Chinese remainder theorem combines these congruences to give $a^k\equiv a(\mathrm{mod}n)$.

RSA chooses $n=pq$ and exponents $e,d$ with $ed\equiv 1(\mathrm{mod}\varphi (n))$. The public key is $(n,e)$, encryption is $c\equiv m^e(\mathrm{mod}n)$, and the secret exponent decrypts via $c^d\equiv m(\mathrm{mod}n)$. Security relies on the difficulty of recovering $\varphi (n)$ from a large unfactored $n$.

Choose a prime with $p^2\mid n$, and set $a=0$, $b=n/p$. Then $b\not\equiv 0(\mathrm{mod}n)$, but $b^2=n^2/p^2$ is divisible by $n$. Therefore $a^k\equiv b^k\equiv 0(\mathrm{mod}n)$ for every $k>1$.

Injective means $f(x)=f(y)$ implies $x=y$. It need not imply surjective on an infinite set: $f(n)=n+1$ is injective from $\mathbb{N}$ to itself but misses its least element.

An element of the left side is $f(s)$ for some $s\in S$ with $f(s)\in T$, exactly an element of $T\cap f(S)$. The second equality need not hold, even for a surjection: map $\{1,2\}$ to a singleton and take $S=\{1\}$, $S^{\prime }=\{2\}$. Then the image of the intersection is empty but the intersection of the images is not.

True. If $p(a,a^{\prime })=p(b,b^{\prime })$, then $a^{\prime }=b^{\prime }$, so $f(a)=g(a^{\prime })=g(b^{\prime })=f(b)$. Injectivity of $f$ gives $a=b$.

True. Given $a^{\prime }\in A^{\prime }$, surjectivity supplies $a\in A$ with $f(a)=g(a^{\prime })$. Then $(a,a^{\prime })$ lies in the fibre product and maps to $a^{\prime }$.

A set is countable when it is finite or admits a bijection with a subset of $\mathbb{N}$. For each degree, the coefficient tuples for polynomials in $A[x]$ form a finite Cartesian power of the countable set $A$; a countable union over degrees is countable. In particular the algebraic numbers, being roots of countably many integer polynomials with finitely many roots each, are countable. Since $\mathbb{R}$ is uncountable, uncountably many real numbers are transcendental.

The set $Q$ is uncountable. Partition $N$ into four-element blocks and, independently on each block, choose either of two fixed-point-free permutations. Binary sequences then inject into $Q$.

Finally choose a line $L$ distinct from and nonparallel to every $L_i$, possible because only countably many directions are excluded. Each $L\cap L_i$ has at most one point, so the union covers only countably many points of the uncountable line $L$. It cannot cover the plane.

With $\omega =B/m$, magnetic force does no work, so $|\dot{x}|$ is constant; its zero $z$ component also makes $\dot{z}$ constant. For the stated data,This is a circle of radius $u/\omega$ centred at $(0,-u/\omega )$.

With $\alpha =\mu /m$, the complex velocity is $\dot{x}+i\dot{y}=u{e}^{-(\alpha +i\omega )t}$. HenceThe path spirals into $u(\alpha ,-\omega )/({\alpha }^2+{\omega }^2)$. As $u$ varies, this lies on $y=-(\omega /\alpha )x$, at distancefrom the origin.

Binet's equation givesConsequently

At infinity the specific energy is $v^2/2$ and $l=bv$, soThe angle through which the velocity is deflected between the incoming and outgoing asymptotes is

Zero energy gives $e=1$. Since the periapsis is $d$, $r_0=2d$ and $l=\sqrt{2kd}$. Put $T=\tan (\theta /2)$; thenTaking $t=0$ at periapsis yields

For axis unit vector $\hat{\omega }$,For a body, replace the sum by $I=\int \rho (x)r_{\perp }^{2}dV$.

For the hollow ball,Energy conservation and $v=R_{+}\Omega$ giveThe factor $\beta$ increases with $\mu$, so if ${\mu }_A>{\mu }_B$, ball B accelerates more and arrives first. Moving the cavity off centre removes material with a larger lever arm and lowers the moment of inertia about the centre of mass; despite the resulting wobble, ball C therefore rolls down faster than A.

With $\gamma =(1-v^2/c^2{)}^{-1/2}$,Differentiation with respect to proper time givesOrthogonality $U\cdot A=0$ implies that in the instantaneous rest frame $A^{\prime \mu }=(0,a^{\prime })$. For collinear motion a Lorentz boost givesFor constant positive proper acceleration $a^{\prime }$,This hyperbola has future null asymptote $x=ct-c^2/a^{\prime }$. A right-moving signal emitted from $x_0$ at $t=0$ follows $x=ct+x_0$; if $x_0<-c^2/a^{\prime }$, it lies beyond that acceleration horizon and never intersects the worldline.