Past Exam Questions

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.

Key idea. There are fewer months than people.

1. Group the 13 people according to their birth month. There are 12 possible groups.

2. If every group contained at most one person, there would be at most 12 people.

3. There are 13 people, so at least one group must contain two or more. Those people were born in the same month, possibly in different years.

Remember. If more objects than boxes are placed in the boxes, some box contains at least two objects.

Key idea. Assume every cage stays below the claimed number.

1. Suppose no cage contains 11 or more rabbits. Then every cage contains at most 10.

2. Across 10 cages, that would allow at most \(10\cdot10=100\) rabbits.

3. But there are 102 rabbits. The supposition is impossible, so some cage contains at least 11.

Remember. Forcing a large group can be proved by calculating the capacity when all groups are smaller.

Key idea. Count how many socks can be present without reaching three of a colour.

1. There are only two colours. Suppose each colour appears at most twice among the socks taken.

2. Then there could be at most \(2+2=4\) socks altogether.

3. We have taken five socks. Therefore one colour must appear at least three times.

Remember. The argument gives a guarantee even in the least favourable selection.

Key idea. Make pairs whose sums are 11.

1. Arrange the available numbers in the five pairs

   \[(1,10),\quad(2,9),\quad(3,8),\quad(4,7),\quad(5,6).\]

2. Each number from 1 to 10 belongs to exactly one of these pairs.

3. If we chose at most one number from each pair, we could choose at most five numbers. Choosing six distinct numbers therefore forces both members of some pair.

4. The two members of that pair have sum 11, as required.

Remember. The useful “boxes” may be pairs that you construct, rather than groups already named in the question.

Key idea. Find the smallest possible total if every cage population is different.

1. Suppose all 14 cages contain different numbers of rabbits. List those numbers from smallest to largest.

2. The first number is at least 0, the second at least 1, the third at least 2, and so on. The fourteenth is at least 13.

3. The total must therefore be at least

   \[0+1+2+\cdots+13=7\cdot13=91.\]

   To get 91, pair the first and last numbers, then the second and second-last: there are seven pairs, each summing to 13.

4. There are only 90 rabbits. So the assumption that every cage population is different cannot be true.

Remember. When numbers must be distinct, ordering them gives a lower bound on their sum.

Key idea. The first coordinate must divide 12.

1. From \(xy=12\), we get \(y=12/x\). For both numbers to be positive integers, \(x\) must be a positive divisor of 12.

2. The positive divisors are \(1,2,3,4,6,12\). Substituting each gives

   \[(1,12),(2,6),(3,4),(4,3),(6,2),(12,1).\]

3. Every listed pair has product 12. There can be no other pair, because all possible positive divisors for the first coordinate have been used.

Remember. For ordered pairs, swapping the coordinates usually gives another answer.

Key idea. First find the possible magnitudes, then restore the signs.

1. Let \(X=|x|\), \(Y=|y|\). Since \(x^2-y^2=15>0\), we have \(X>Y\ge0\), and

   \[(X-Y)(X+Y)=15.\]

2. The positive factor pairs of 15, with the smaller first, are \((1,15)\) and \((3,5)\). If \(u=X-Y\), \(v=X+Y\), then

   \[X=\frac{u+v}{2},\qquad Y=\frac{v-u}{2}.\]

3. The pair \((1,15)\) gives \((X,Y)=(8,7)\); the pair \((3,5)\) gives \((X,Y)=(4,1)\).

4. Squares do not depend on signs. Each magnitude pair gives four solutions:

   \[\begin{gathered}(8,7),(8,-7),(-8,7),(-8,-7),\\(4,1),(4,-1),(-4,1),(-4,-1).\end{gathered}\]

5. Both magnitudes work because \(64-49=15\) and \(16-1=15\). Every factor pair was considered, so the list is complete.

Remember. For equations involving squares, separating magnitudes from signs can make completeness easier to check.

Key idea. Factor the difference of squares, then check the reconstruction carefully.

1. Let \(X=|x|\), \(Y=|y|\). We have \(X>2Y\), so the two factors in

   \[(X-2Y)(X+2Y)=12\]

   are positive.

2. Set \(u=X-2Y\), \(v=X+2Y\). Then

   \[X=\frac{u+v}{2},\qquad Y=\frac{v-u}{4}.\]

   Both quantities must be integers.

3. Check every positive factor pair of 12:

   \(u\)	\(v\)	\(X\)	\(Y\)
   \(1\)	\(12\)	\(\frac{13}{2}\)	\(\frac{11}{4}\)
   \(2\)	\(6\)	\(4\)	\(1\)
   \(3\)	\(4\)	\(\frac{7}{2}\)	\(\frac{1}{4}\)

   Only the middle row gives integers.

4. Thus \(|x|=4\), \(|y|=1\). Restoring both signs gives \((4,1),(4,-1),(-4,1),(-4,-1)\). Each satisfies \(16-4=12\).

Remember. The reconstruction formula determines the divisibility conditions. Here the factor difference must be divisible by 4.

Key idea. Relate the roots to the coefficients by expanding a product.

1. If the two integer roots are \(r,s\), then

   \[x^2-ax+12=(x-r)(x-s)=x^2-(r+s)x+rs.\]

2. Comparing coefficients gives \(rs=12\) and \(a=r+s\). Since the product is positive, the roots have the same sign.

3. The positive unordered pairs are \((1,12),(2,6),(3,4)\), with sums 13, 8 and 7. The negative pairs have sums \(-13,-8,-7\).

4. Conversely, each of those six sums gives a factorisation with the listed integer roots. All integer factor pairs of 12 are covered.

Remember. Comparing coefficients gives both a necessary condition and a way to construct the roots.

Key idea. Complete squares so that the difference has a fixed value.

1. Write the required product as \(m(m+1)\), where \(m\) is a positive integer. Our equation is \(n^2+2n+12=m(m+1)\).

2. Multiply by 4 and complete squares:

   \[(2m+1)^2-[2(n+1)]^2=45.\]

3. Let \(u=2m+1-2(n+1)\), \(v=2m+1+2(n+1)\). Then \(uv=45\). The second factor is positive, so the first is positive too; also \(u<v\). Thus the possible pairs are \((1,45),(3,15),(5,9)\).

4. Add and subtract the definitions to recover

   \[m=\frac{u+v-2}{4},\qquad n=\frac{v-u}{4}-1.\]

   This gives

   \(u\)	\(v\)	\(m\)	\(n\)
   \(1\)	\(45\)	\(11\)	\(10\)
   \(3\)	\(15\)	\(4\)	\(2\)
   \(5\)	\(9\)	\(3\)	\(0\)

5. Exclude \(n=0\), because n must be positive. For \(n=2\), the expression is \(20=4\cdot5\); for \(n=10\), it is \(132=11\cdot12\). Both remaining cases work.

Remember. A finite list of divisor pairs can replace a search over infinitely many integers.

Key idea. Treat the repeated expression as a single unknown.

1. Put \(u=x^2-3x\). The equation becomes

   \[u^2-2u-8=(u-4)(u+2)=0,\]

   so \(u=4\) or \(u=-2\).

2. If \(u=4\), then \(x^2-3x-4=0\), or \((x-4)(x+1)=0\). Thus \(x=4\) or \(x=-1\).

3. If \(u=-2\), then \(x^2-3x+2=0\), or \((x-1)(x-2)=0\). Thus \(x=1\) or \(x=2\).

4. Each branch came from an equivalent factorisation and both branches were solved completely. All four listed values therefore satisfy the original equation, and there are no others.

Remember. A substitution is complete only after every possible value of the new variable has been converted back.

Key idea. Subtract the two equations to split the system into two cases.

1. Subtraction gives

   \[x^2-y^2+y-x=(x-y)(x+y-1)=0.\]

   Therefore either \(x=y\) or \(x+y=1\).

2. If \(x=y\), then \(x^2+x=12\), so \((x-3)(x+4)=0\). This gives \((3,3)\) and \((-4,-4)\), both of which satisfy the system.

3. If \(x+y=1\), substitute \(y=1-x\):

   \[x^2-x-11=0,\qquad x=\frac{1\pm\sqrt{45}}2=\frac{1\pm3\sqrt5}2.\]

   The value of y uses the opposite sign.

4. The two values in the last case are the roots of \(t^2-t-11=0\) and have sum 1. Hence \(x^2=x+11\) and \(y^2=y+11\), giving \(x^2+y=y^2+x=12\). So these pairs work too.

5. The factored difference forces one of our two cases, so the four pairs found form the complete solution set.

Remember. Subtracting symmetric equations can reveal cases that direct substitution hides.

Key idea. Clear the fractions, then complete a product.

1. Both reciprocals are positive, so each is less than \(1/6\). Thus \(x>6\) and \(y>6\).

2. Multiply the equation by \(6xy\): \(6x+6y=xy\). Rearranging and adding 36 gives

   \[xy-6x-6y+36=36,\qquad (x-6)(y-6)=36.\]

3. Set \(u=x-6\), \(v=y-6\). These are positive integers with product 36. The positive divisors of 36 are \(1,2,3,4,6,9,12,18,36\). Each possible u determines \(v=36/u\).

4. Adding 6 to both coordinates gives

   \[\begin{gathered}(7,42),(8,24),(9,18),(10,15),(12,12),\\(15,10),(18,9),(24,8),(42,7).\end{gathered}\]

5. Every positive factor pair was included. Reversing the algebra recovers the original equation, so every listed pair works and there are no others.

Remember. Adding a constant can turn a two-variable equation into a fixed product with finitely many factor pairs.

Key idea. Replace the two variables by their sum and product, then bound the sum.

1. Let \(s=a+b\), \(p=ab\). The cube identity gives

   \[a^3+b^3=s^3-3ps=s^2.\]

   Since \(s>0\), division by s yields \(p=s(s-1)/3\).

2. The square \((a-b)^2\ge0\) says \(4p\le s^2\). Substitute the expression for p and divide by positive s:

   \[\frac{4s(s-1)}3\le s^2\quad\Longrightarrow\quad4(s-1)\le3s\quad\Longrightarrow\quad s\le4.\]

3. Positive integers a and b give integer \(s\ge2\). Only \(s=2,3,4\) remain. For \(s=2\), we get \(p=2/3\), impossible for an integer product.

4. For \(s=3\), we get \(p=2\), giving \((a,b)=(1,2)\) or \((2,1)\). For \(s=4\), we get \(p=4\), giving \((a,b)=(2,2)\).

5. Check: \(1^3+2^3=9=(1+2)^2\), and \(2^3+2^3=16=(2+2)^2\). All possibilities for the sum have been considered.

Remember. An inequality can reduce an integer equation to a short, complete list of cases.

Key idea. Express the square of the prime as a product using the given equation.

1. Write \(p=a-b\), which is a positive prime, and \(s=a+b\). Then

   \[p^2=(a-b)^2=s^2-4ab.\]

2. The hypothesis gives \(ab=3c^2-cs\). Substitute and factor:

   \[p^2=s^2+4cs-12c^2=(s-2c)(s+6c).\]

3. The second factor is positive. Since their product is positive, the first is positive too. They are distinct, because their difference is \(8c>0\).

4. The positive factor pairs of a prime square \(p^2\), in increasing order, are \((1,p^2)\) and \((p,p)\). Our factors are distinct, so

   \[s-2c=1,\qquad s+6c=p^2.\]

5. Subtract these two equations: \(8c=p^2-1\). Hence \(8c+1=p^2\), the square of the prime \(a-b\).

Remember. A prime square has very few factor pairs. Positivity and the order of the factors decide which pair is possible.

Key idea. Leave a multiple of 3 after each of your turns.

1. The first player takes 2 stones, leaving 12.

2. After that, respond to an opponent taking 1 by taking 2; respond to an opponent taking 2 by taking 1. The two moves always remove 3 stones in total.

3. After each reply, the pile size is 9, then 6, then 3, then 0. These are all multiples of 3.

4. The reply is always legal. When a positive multiple of 3 is on the table, the opponent can take only 1 or 2, leaving enough for the matching reply.

5. In particular, from a pile of 3 the opponent leaves 1 or 2, which the first player takes. The first player gets the last stone.

Remember. A strategy explains how to answer every legal move, not just one possible game.

Key idea. Make each opponent-move-and-reply remove four stones.

1. The first player takes 2 stones from 30, leaving 28, a multiple of 4.

2. Then use these replies: if the opponent takes 1, take 3; if the opponent takes 2, take 2; if the opponent takes 3, take 1.

3. After each completed pair of moves, the remaining pile is a smaller multiple of 4: 24, 20, 16, 12, 8, 4, then 0.

4. Every reply is allowed and enough stones remain for it. From 4 stones, whatever the opponent takes, the first player can take the rest.

5. The opponent is then left with no move, so loses.

Remember. The same strategy changes naturally when the allowed moves change from 1-2 to 1-3.

Key idea. The second player can restore equal piles after every move.

1. The piles start equal: 10 stones each.

2. Whenever the first player removes a number of stones from one pile, the second player removes the same number from the other pile.

3. This reply is legal because the piles were equal just before the first player’s move. Afterwards they are equal again, so the rule can be repeated.

4. Eventually the first player empties a pile. The second player empties the equally large other pile and takes the last stone.

Remember. Symmetry is useful when you can restore it after every opponent move.