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(a) Find all prime numbers p such that $$4p^2+1$$ and $$6p^2+1$$ are also primes.
(b) Determine real numbers x, y, z, u such that
$$xyz+xy+yz+zx+x+y+z=7$$,
$$yzu+yz+zu+uy+y+z+u=9$$,
$$zux+zu+ux+xz+z+u+x=9$$,
$$uxy+ux+xy+yu+u+x+y=9$$.
For part (a), checking small primes shows $$p=5$$ gives $$4(5)^2+1=101$$ and $$6(5)^2+1=151$$, both prime, while any prime $$p>5$$ must be of the form $$5m+1$$ or $$5m+4$$, and in either case one of $$4p^2+1$$ or $$6p^2+1$$ becomes a multiple of 5 greater than 5, hence not prime. So $$p=5$$ is the only solution. For part (b), each equation factors nicely, e.g. the first as $$(z+1)(x+1)(y+1)=8$$, and similarly $$(u+1)(y+1)(z+1)=10$$, $$(x+1)(z+1)(u+1)=10$$, $$(y+1)(u+1)(x+1)=10$$; multiplying all four and taking a fourth root gives $$(x+1)(y+1)(z+1)(u+1)=20$$, and dividing this by each individual equation gives $$u+1=2$$, $$x+1=2$$, $$y+1=2$$, $$z+1=2$$, so $$x=y=z=1$$ and $$u=2$$.
If x, y, z, p, q, r are distinct real numbers such that
$$\left(\frac{1}{x+p}+\frac{1}{y+p}+\frac{1}{z+p}\right) = \frac{1}{p}$$
$$\left(\frac{1}{x+q}+\frac{1}{y+q}+\frac{1}{z+q}\right) = \frac{1}{q}$$
$$\left(\frac{1}{x+r}+\frac{1}{y+r}+\frac{1}{z+r}\right) = \frac{1}{r}$$,
find the numerical value of $$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$$.
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ADC and ABC are triangles such that $$AD=DC$$ and $$CA=AB$$. If $$\angle CAB=20^\circ$$ and $$\angle ADC=100^\circ$$, without using trigonometry, prove that $$AB=BC+CD$$.
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(a) a, b, c, d are positive real numbers such that $$abcd=1$$.
Prove that $$\frac{1+ab}{a+1}+\frac{1+bc}{b+1}+\frac{1+cd}{c+1}+\frac{1+da}{d+1} \ge 4$$.
(b) In a scalene triangle ABC, $$\angle BAC=120^\circ$$. The bisectors of angles A, B and C meet the opposite sides in P, Q and R respectively. Prove that the circle on QR as diameter passes through the point P.
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(a) Prove that $$x^4+3x^3+6x^2+9x+12$$ cannot be expressed as a product of two polynomials of degree 2 with integer coefficients. (b) $$2n+1$$ segments are marked on a line. Each of these segments intersects at least n other segments. Prove that one of these segments intersects all other segments.
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If a, b, c, d are positive real numbers such that $$a^2+b^2=c^2+d^2$$ and $$a^2+d^2-ad = b^2+c^2+bc$$, find the value of $$\frac{ad+bc}{ab+cd}$$.
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