Proof & Number Theory

Part (i).

Differentiate $y=27x^3-27x^2+4$ to get $\frac{dy}{dx}=81x^2-54x=27x(3x-2)$. Setting this to zero gives $x=0$ or $x=\frac{2}{3}$.

- At $x=0$: $y=4$, so the turning point is $(0,4)$.
- At $x=\frac{2}{3}$: $y=27\cdot \frac{8}{27}-27\cdot \frac{4}{9}+4=8-12+4=0$, so the turning point is $\left(\frac{2}{3},0\right)$.

Since the leading coefficient is positive, this is a cubic that rises to the right. The $y$-intercept is $4$. For the $x$-intercepts, set $y=0$: the fact that $x=\frac{2}{3}$ is a double root (a turning point touching the axis) means $27x^3-27x^2+4=27(x+\frac{1}{3})(x-\frac{2}{3}{)}^2$, giving a third crossing at $x=-\frac{1}{3}$. Sketch the curve accordingly: it crosses the $x$-axis at $-\frac{1}{3}$, has a local max at $(0,4)$, and touches zero at $\left(\frac{2}{3},0\right)$.

Deducing the inequality. For $x⩾0$, the sketch shows $y⩾0$, i.e.$$27x^3-27x^2+4⩾0.$$Rearrange: $4⩾27x^2-27x^3=27x^2(1-x)$, which gives $x^2(1-x)⩽\frac{4}{27}$ for all $x⩾0$.

Proof by contradiction (the three-variable part). Suppose for contradiction that all three of $bc(1-a)$, $ca(1-b)$, $ab(1-c)$ exceed $\frac{4}{27}$. Multiply these three inequalities together:$$bc(1-a)\cdot ca(1-b)\cdot ab(1-c)>{\left(\frac{4}{27}\right)}^3.$$The left-hand side factors as $a^2(1-a)\cdot b^2(1-b)\cdot c^2(1-c)$. Since $a,b,c\in (0,1)\subset [0,\infty )$, the inequality proved above applies to each factor:$$a^2(1-a)⩽\frac{4}{27},b^2(1-b)⩽\frac{4}{27},c^2(1-c)⩽\frac{4}{27}.$$Multiplying: $a^2(1-a)\cdot b^2(1-b)\cdot c^2(1-c)⩽{\left(\frac{4}{27}\right)}^3$. This contradicts the strict inequality above. Hence at least one of the three numbers is at most $\frac{4}{27}$.

Part (ii).

The key one-variable bound here is $x(1-x)⩽\frac{1}{4}$ for all $x$, which follows from completing the square: $x(1-x)=\frac{1}{4}-{\left(x-\frac{1}{2}\right)}^2⩽\frac{1}{4}$, with equality at $x=\frac{1}{2}$.

Suppose for contradiction that both $p(1-q)>\frac{1}{4}$ and $q(1-p)>\frac{1}{4}$. Multiply:$$p(1-q)\cdot q(1-p)>\frac{1}{16}.$$But the left-hand side is $p(1-p)\cdot q(1-q)$. Since $p,q\in (0,1)$, the bound above gives $p(1-p)⩽\frac{1}{4}$ and $q(1-q)⩽\frac{1}{4}$, so $p(1-p)\cdot q(1-q)⩽\frac{1}{16}$. This is a contradiction. Hence at least one of $p(1-q)$, $q(1-p)$ is at most $\frac{1}{4}$.

Part (i). By the definition of the Laplace transform,$$\mathcal{L}\{e^{-bt}f(t)\}(s)={\int }_0^{\infty }{e}^{-st}{e}^{-bt}f(t)dt={\int }_0^{\infty }{e}^{-(s+b)t}f(t)dt=F(s+b).$$The exponential $e^{-bt}$ simply shifts the parameter from $s$ to $s+b$ inside the integral.

Part (ii). Set $u=at$, so $du=adt$ and when $t=0$ we have $u=0$, when $t\to \infty$ we have $u\to \infty$. Then$$\mathcal{L}\{f(at)\}(s)={\int }_0^{\infty }{e}^{-st}f(at)dt={\int }_0^{\infty }{e}^{-s(u/a)}f(u)\frac{du}{a}=\frac{1}{a}{\int }_0^{\infty }{e}^{-(s/a)u}f(u)du=\frac{1}{a}F\left(\frac{s}{a}\right).$$Scaling time by $a$ compresses or stretches the signal; the factor $a^{-1}$ corrects for the change in the measure.

Part (iii). Integrate by parts with $u=e^{-st}$ and $dv=f^{\prime }(t)dt$:$$\mathcal{L}\{f^{\prime }(t)\}(s)={\int }_0^{\infty }{e}^{-st}{f}^{\prime }(t)dt=[e^{-st}f(t){]}_0^{\infty }+s{\int }_0^{\infty }{e}^{-st}f(t)dt.$$For suitable $f$, the boundary term at infinity vanishes (the decay of $e^{-st}$ dominates), so the boundary term equals $0-f(0)$. The remaining integral is $F(s)$, giving$$\mathcal{L}\{f^{\prime }(t)\}=sF(s)-f(0).$$Intuitively, differentiation in the time domain becomes multiplication by $s$ in the transform domain (up to the initial value).

Part (iv). Take $f(t)=\sin t$. Note $f(0)=0$, and $f^{\prime }(t)=\cos t$, $f^{\prime \prime }(t)=-\sin t=-f(t)$. Applying result (iii) twice:$$\mathcal{L}\{\cos t\}=sF(s)-f(0)=sF(s),$$$$\mathcal{L}\{-\sin t\}=s\mathcal{L}\{\cos t\}-\cos (0)=s^{2}F(s)-1.$$But $\mathcal{L}\{-\sin t\}=-F(s)$, so$$-F(s)=s^{2}F(s)-1⟹F(s)(1+s^2)=1⟹F(s)=\frac{1}{s^2+1}.$$Finding the Laplace transform of $e^{-pt}\cos (qt)$. First, use result (ii) to find the transform of $\cos (qt)$ from that of $\sin t$.

Apply (iii) to $g(t)=\cos (qt)$: note $g(0)=1$ and $g^{\prime }(t)=-q\sin (qt)$, so$$\mathcal{L}\{-q\sin (qt)\}=sG(s)-1,$$where $G(s)=\mathcal{L}\{\cos (qt)\}$. By result (ii) with $a=q$ applied to $f(t)=\sin t$:$$\mathcal{L}\{\sin (qt)\}=\frac{1}{q}F\left(\frac{s}{q}\right)=\frac{1}{q}\cdot \frac{1}{(s/q{)}^2+1}=\frac{q}{s^2+q^2}.$$Therefore$$-q\cdot \frac{q}{s^2+q^2}=sG(s)-1⟹G(s)=\frac{s}{s^2+q^2}.$$Finally, apply result (i) with $b=p$ to $g(t)=\cos (qt)$:$$\mathcal{L}\{e^{-pt}\cos (qt)\}=G(s+p)=\frac{s+p}{(s+p{)}^2+q^2}.$$

Part (i). By inspection, $x=1$, $y=0$ satisfies $x^2-2y^2=1^2-0=1$.

Now suppose $x=p$, $y=q$ is a solution. Set $x=3p+4q$, $y=2p+3q$ and compute:$$x^2-2y^2=(3p+4q{)}^2-2(2p+3q{)}^2=9p^2+24pq+16q^2-2(4p^2+12pq+9q^2)$$$$=9p^2+24pq+16q^2-8p^2-24pq-18q^2=p^2-2q^2=1.$$So $(3p+4q,2p+3q)$ is also a solution whenever $(p,q)$ is.

Starting from $(1,0)$: apply once to get $(3,2)$; apply again to get $(3\cdot 3+4\cdot 2,2\cdot 3+3\cdot 2)=(17,12)$. Both have $x$ odd and $y$ even. Check: $17^2-2\cdot 12^2=289-288=1$. ✓

Part (ii). Write $x=2n-1$ (odd) and $y=2m$ (even) for integers $m$ and $n$. Substituting into $(\ast )$:$$(2n-1{)}^2-2(2m{)}^2=1$$$$4n^2-4n+1-8m^2=1$$$$4n^2-4n=8m^2$$$$n^2-n=2m^2$$$$n(n-1)=2m^2.$$Since $n(n-1)$ is the product of two consecutive integers, we can write $n(n-1)=\frac{1}{2}\cdot 2n(2n-2)$... more directly: set $x=2n-1$, $y=2m$, then $n^2=\frac{1}{2}m(m+1)$... Let us redo cleanly. Write $x=2n+1$, $y=2m$. Then:$$(2n+1{)}^2-2(2m{)}^2=1⟹4n^2+4n+1-8m^2=1⟹4n^2+4n=8m^2⟹n^2+n=2m^2.$$so $m^2=\frac{1}{2}n(n+1)=\frac{1}{2}{m}^{\prime }(m^{\prime }+1)$ with $m^{\prime }=n$. Renaming $m^{\prime }\to m$ and $n\to n$ gives $n^2=\frac{1}{2}m(m+1)$ only after swapping roles: set $m=2n$, $n\mapsto m$. The cleanest path: let $x=2n-1$, $y=2m$. Then $(\ast )$ gives $n(n-1)=2m^2$, i.e. $m^2=\frac{1}{2}(n-1)n=\frac{1}{2}k(k+1)$ where $k=n-1$. Relabelling $k\to m$ and $m\to n$ yields $n^2=\frac{1}{2}m(m+1)$ as required.

Part (iii). We have $b^3=c^4-a^2=(c^2-a)(c^2+a)$. Since $b$ is prime, $b^3$ factors as $1\cdot b^3$ or $b\cdot b^2$. Let $c^2-a=s$ and $c^2+a=t$ with $st=b^3$ and $t>s>0$, $t+s=2c^2$, $t-s=2a$.

Case 1: $s=1$, $t=b^3$. Then $2c^2=b^3+1$ and $2a=b^3-1$, giving $a=\frac{b^3-1}{2}$ and $c^2=\frac{b^3+1}{2}$.

Case 2: $s=b$, $t=b^2$. Then $2c^2=b^2+b=b(b+1)$ and $2a=b^2-b=b(b-1)$, giving $a=\frac{b(b-1)}{2}$ and $c^2=\frac{b(b+1)}{2}$.

Notice that $c^2=\frac{b(b+1)}{2}$ is precisely the triangular-number condition from part (ii) (with $n=c$, $m=b$ or similar). The solution $(x,y)=(17,12)$ from part (i) gives $n^2=\frac{1}{2}m(m+1)$ with $c=12$, $b=\dots$ — we need $c^2=144$ and $b(b+1)=288$, so $b=16$ (since $16\times 17=272$... try $b=17$: $17\times 18=306$). Instead use Case 1 directly: take $b=2$ (prime). Then $a=\frac{7}{2}$ (not integer). Try $b=3$: $a=13$, $c^2=14$ (not a perfect square). Try $b=7$: $a=171$, $c^2=172$ (no). Case 2 with $b=3$: $a=3$, $c^2=6$ (no). With $b=7$: $a=21$, $c^2=28$ (no). With $b=2$: $a=1$, $c^2=3$ (no).

For $b$ not prime, we need $a^2+b^3=c^4$ with $a,b,c$ positive integers. Take $b=2^2=4$ (not prime). We want $c^4-a^2=64$, so $(c^2-a)(c^2+a)=64$. Try $c^2-a=2$, $c^2+a=32$: then $c^2=17$ (no). Try $c^2-a=4$, $c^2+a=16$: $c^2=10$ (no). Try $c^2-a=8$, $c^2+a=8$: $a=0$ (not positive). Try $b=8$: $b^3=512$. $(c^2-a)(c^2+a)=512$. Take $c^2-a=8,c^2+a=64$: $c^2=36$, so $c=6$, $a=28$. Check: $28^2+8^3=784+512=1296=6^4$. ✓ So $\mathbf{a}=28,\mathbf{b}=8,\mathbf{c}=6$.