Key idea. Track whether the number is odd or even.

1. The starting number, 4, is even.

2. Adding 2 to an even number gives another even number. Repeating this move any number of times therefore always gives an even number.

3. The target 99 is odd, so it can never appear.

Remember. A property preserved by every move is called an invariant. Here it is evenness.

Key idea. Moving a stone changes its location, not the total number of stones.

1. The original piles contain \(3+5+7=15\) stones altogether.

2. Moving one stone out of a pile removes 1 there but adds 1 to another pile. The total stays 15 after every move.

3. The proposed final piles would contain \(4+6+10=20\) stones. Since \(20\ne15\), they cannot be reached.

Remember. Check a simple total before experimenting with many moves.

Key idea. Count tails and examine every kind of move.

1. Initially there are no tails, so the tail count is 0, an even number.

2. If we flip two heads, the tail count increases by 2. If we flip two tails, it decreases by 2. If we flip one head and one tail, the count does not change.

3. In all three cases its parity stays the same. Thus the number of tails remains even after every move.

4. Having all nine coins tails up would give nine tails, which is odd. That is impossible.

Remember. Checking all types of move explains why an invariant really is preserved.

Key idea. Count black and white squares, not just the total area.

1. An \(8\times8\) chessboard has 32 black and 32 white squares.

2. Two opposite corners have the same colour. Removing them leaves 30 squares of that colour and 32 of the other.

3. Each domino covers two squares sharing a side. On a chessboard these always have different colours, so every domino covers one black and one white square.

4. Any collection of dominoes therefore covers equal numbers of the two colours. It cannot cover a region with 30 of one and 32 of the other.

Remember. An even area is necessary for a domino tiling, but colour counts can reveal another obstruction.

Key idea. Use parity for the even-sized moves; give a construction for five-cup moves.

1. Initially there are 0 upright cups. If a move flips \(k\) cups, of which \(u\) were upright, the upright count changes by \(k-2u\). When \(k\) is even, this change is even.

2. So with either two-cup moves or six-cup moves, the number of upright cups stays even. It cannot become 33. Parts (a) and (b) are impossible.

3. For part (c), put the 33 cups in a circle and number them 1 to 33. Make one move starting at each cup: flip that cup and the next four clockwise.

4. The first moves flip 1-5, then 2-6, and so on. Near the end, wrap around the circle: for example, the move starting at 30 flips 30, 31, 32, 33 and 1.

5. Every cup is in exactly five of these blocks. For example, cup 1 is flipped in the moves starting at 1, 30, 31, 32 and 33. The same count holds for every cup.

6. Each cup is therefore flipped five times. Starting upside down, it ends upright after an odd number of flips. After all 33 moves, every cup is upright.

Remember. A “no” answer needs an obstruction; a “yes” answer needs a method that actually works.

Key idea. Group terms to create a repeated bracket.

1. Group the first two terms and the last two:

   \[(ab+ac)+(bd+cd).\]

2. Take out the common factor in each group:

   \[a(b+c)+d(b+c).\]

3. The whole bracket \(b+c\) is now a common factor, giving \((a+d)(b+c)\).

Remember. Grouping works when it produces the same bracket more than once.

Key idea. Remove the common factor first, then factor the remaining quadratic.

1. Every term contains \(3xy\):

   \[6x^3y-9x^2y^2+3xy^3=3xy(2x^2-3xy+y^2).\]

2. To factor the bracket, split its middle term:

   \[2x^2-3xy+y^2=2x^2-2xy-xy+y^2.\]

3. Group and factor:

   \[2x(x-y)-y(x-y)=(2x-y)(x-y).\]

4. Restore the common factor \(3xy\). The complete factorisation is \(3xy(2x-y)(x-y)\).

Remember. Factoring can take more than one step. Do not stop after taking out the common factor.

Key idea. Add and subtract the same term to reveal a square.

1. Compare with \((x^2+1)^2=x^4+2x^2+1\). Our expression has only \(x^2\) in the middle, so write

   \[x^4+x^2+1=(x^2+1)^2-x^2.\]

2. This is a difference of squares:

   \[\bigl((x^2+1)-x\bigr)\bigl((x^2+1)+x\bigr).\]

3. Rearranging the terms in each bracket gives \((x^2-x+1)(x^2+x+1)\).

Remember. Adding and subtracting the same quantity changes the form without changing the value.

Key idea. A composite number can be written as two integer factors greater than 1.

1. Treat the expression as a quadratic in \(n^2\). The numbers 1 and 3 have sum 4 and product 3, so

   \[n^4+4n^2+3=(n^2+1)(n^2+3).\]

2. Because \(n\) is a positive integer, \(n\ge1\). Thus \(n^2+1\ge2\) and \(n^2+3\ge4\).

3. Both factors are integers greater than 1. Their product is therefore composite for every allowed value of n.

Remember. Writing a product is not enough: check that neither factor can be 1.

Key idea. Pair the outside numbers and the inside numbers.

1. Let \(u=2020\cdot2026\). The middle product differs from it by 8, because

   \[\begin{aligned}(2020+2)(2020+4)&=2020^2+6\cdot2020+8,\\2020(2020+6)&=2020^2+6\cdot2020.\end{aligned}\]

2. Therefore \(2022\cdot2024=u+8\), and the whole expression becomes \(u(u+8)+16\).

3. Now complete the square:

   \[u(u+8)+16=u^2+8u+16=(u+4)^2.\]

4. Since \(u+4=2020\cdot2026+4=4\,092\,524\), we have even found its integer square root.

Remember. Large numbers can hide a short identity. Compare the products before multiplying everything.

Key idea. Recognise both terms as perfect cubes.

1. Write \(8a^3=(2a)^3\) and \(27b^3=(3b)^3\).

2. The sum-of-cubes factorisation is

   \[u^3+v^3=(u+v)(u^2-uv+v^2).\]

3. Substitute \(u=2a\), \(v=3b\):

   \[8a^3+27b^3=(2a+3b)(4a^2-6ab+9b^2).\]

Remember. In the quadratic factor for a sum of cubes, the middle sign is negative.

Key idea. Use a difference of squares, followed by two cube identities.

1. First,

   \[x^6-1=(x^3-1)(x^3+1).\]

2. Factor each cubic:

   \[\begin{aligned}x^3-1&=(x-1)(x^2+x+1),\\x^3+1&=(x+1)(x^2-x+1).\end{aligned}\]

3. Combining the factors gives

   \[x^6-1=(x-1)(x+1)(x^2+x+1)(x^2-x+1).\]

4. Each quadratic has discriminant \(1-4=-3\), so neither has real roots. In particular, neither splits into linear factors with integer coefficients.

Remember. After one factorisation, inspect each remaining factor for another useful identity.

Key idea. Create a square by adding and subtracting the missing middle term.

1. We know \((x^2+2y^2)^2=x^4+4x^2y^2+4y^4\). Therefore

   \[x^4+4y^4=(x^2+2y^2)^2-4x^2y^2.\]

2. The subtracted term is also a square: \(4x^2y^2=(2xy)^2\).

3. Apply the difference-of-squares identity:

   \[x^4+4y^4=(x^2-2xy+2y^2)(x^2+2xy+2y^2).\]

Remember. A sum can sometimes be turned into a difference of squares. This identity is often called Sophie Germain’s identity.

Key idea. Use C2.3 and ask when one of the integer factors can equal 1.

1. The factorisation is

   \[x^4+4y^4=\bigl((x-y)^2+y^2\bigr)\bigl((x+y)^2+y^2\bigr).\]

2. For positive integers x and y, both factors are positive integers. The second is at least \(2^2+1^2=5\).

3. If their product is prime, the first factor must therefore be 1:

   \[(x-y)^2+y^2=1.\]

   Since \(y\ge1\), we must have \(y=1\) and \(x-y=0\), giving \(x=1\).

4. Conversely, \((x,y)=(1,1)\) gives \(1+4=5\), which is prime. No other positive integer pair can work.

Remember. A prime written as a product of positive integers forces all but one factor to be 1.

Key idea. Factor the polynomial and isolate the case in which a factor is 1.

1. Try a factor \(n^2+n+1\). Dividing \(n^5+n+1\) by it starts with \(n^3\), then \(-n^2\), then 1, and gives

   \[n^5+n+1=(n^2+n+1)(n^3-n^2+1).\]

2. You can check the identity by multiplying the brackets: the coefficients of \(n^4,n^3,n^2\) cancel, leaving \(n^5+n+1\).

3. If \(n=1\), the expression equals 3, which is prime.

4. If \(n\ge2\), then \(n^2+n+1\ge7\) and

   \[n^3-n^2+1=n^2(n-1)+1\ge5.\]

   Both factors exceed 1, so the expression is composite. These cases cover every positive integer n.

Remember. A polynomial factorisation turns a primality problem into a check of small exceptional cases.