MO Landesrunde Klasse 9 Mock Paper 2: Questions, Hints and Solutions

Direct expansion shows that both the forward and inverse transformations preserve x²−2y². Forward iteration from (1,0) gives (3,2) and then positive increasing integer pairs, so every listed pair works. Conversely let x,y>0 solve the equation. y=1 gives x²=3, impossible; y=2 gives x=3. For y>2, we have $\sqrt2\lt x/y\lt 3/2$, since x²/y²=2+1/y²<9/4. Thus $x^{\prime}=3x-4y\gt 0$ and $y^{\prime}=3y-2x\gt 0$, while x>y implies $y^{\prime}\lt y$. The inverse pair is therefore a smaller positive solution. Repeating must reach (3,2), whose inverse is (1,0). Reversing the steps proves that every positive solution occurs in the stated sequence.

Conclusion: Exactly the positive pairs generated from (1,0) by the given recurrence.

Review the idea: Factorisation integer solutions · Proof methods

Suppose such a polynomial P existed and let P(1)=p. Integer coefficients imply $P(1+kp)\equiv P(1)\equiv0\pmod p$ for every positive integer k. By assumption each of these values is a positive prime, so it must equal p. Therefore the polynomial P(x)−p has infinitely many distinct roots 1+p,1+2p,… . A nonzero polynomial of degree d has at most d roots, so P(x)−p must be identically zero. This makes P constant, contradicting the hypothesis. No growth or sign assumption is needed beyond the stated primality.

Conclusion: No nonconstant integer-coefficient polynomial has that property.

Review the idea: Remainder and factor theorems · Polynomial roots and multiplicity

Use directed angles modulo 180 degrees so the proof also covers intersections on extensions. In circle ABCE, $\angle ACE\equiv\angle ABE$, since both subtend chord AE. In circle ABDF, $\angle ADF\equiv\angle ABF$, since both subtend chord AF. As B,E,F are collinear, the two angles at B are equal modulo 180 degrees. Therefore the line angles from AC to CE and from AD to DF agree. The lines AC and AD are the same line, so CE and DF are parallel. The distinctness assumptions ensure all referenced chords and angles are defined.

Conclusion: CE is parallel to DF.

Review the idea: Cyclic and tangential quadrilaterals · Angles

We claim the losing positions are exactly n≡0 or 2 modulo 7. A legal move subtracting 1,3,4 from residue 0 reaches 6,4,3, and from residue 2 reaches 1,6,5, never 0 or 2. Every other residue has a move to 0 or 2: from residues 1,3,4,5,6 subtract 1,1,4,3,4 respectively. These moves are legal even at the smallest positive representatives. Strong induction on n now proves the classification: a candidate losing position has only winning followers, while any other position has a move to a smaller losing one. The base n=0 is as specified.

Conclusion: Exactly n≡0 or 2 modulo 7.

Review the idea: Strong induction · Remainders

The number b must be odd because $b^2=2^a+1$ is odd, so b≥3. Thus $(b-1)(b+1)=2^a$ is a product of positive powers of 2 differing by 2. Their gcd is 2. Consequently the smaller factor is exactly 2: otherwise both would be divisible by 4 and could not differ by 2. Hence b−1=2 and b+1=4, giving b=3 and a=3. Direct substitution verifies $2^3+1=9$.

Conclusion: Only (a,b)=(3,3).

Review the idea: Unique prime factorisation · Greatest common divisor

Starting from u₁=1, every next term is a positive real number greater than its predecessor, so all divisions are valid and $u_n$≥1. Squaring the recurrence gives $u_{n+1}^2-u_n^2=2+1/u_n^2$. Therefore each such increment lies between 2 and 3. Summing the first n−1 increments gives $1+2(n-1)\le u_n^2\le1+3(n-1)$. Taking nonnegative square roots gives the required bounds. For n=1 both sides agree with the initial value.

Conclusion: The bounds hold for all n≥1; all terms are positive.

Review the idea: Sequences and sums · Nonlinear recurrences and substitutions