Key idea. A question asking whether something can happen is settled by one working example.

1. Try the three positive integers 1, 2 and 3. Their sum is \(1+2+3=6\).

2. Their product is \(1\cdot2\cdot3=6\), so it is equal to their sum.

3. All three numbers are positive integers, as required. We have checked every condition of the example.

Remember. One example proves that something is possible. We do not need to list every possible example.

Key idea. Read “Every champion does well” in the direction it is stated.

1. The mother’s statement tells us something about champions: they all belong to the group of pupils who do well at school.

2. It does not say that champions are the only pupils in that group. Someone may do well without being a champion.

3. Imagine Kolya is a pupil who does well and is not a champion. This agrees with his mother’s statement but makes his own conclusion false. Therefore the conclusion does not follow.

Remember. “Every A is B” does not automatically mean “Every B is A”.

Key idea. Look for two divisors that share a factor.

1. Choose \(n=12\), \(a=4\) and \(b=6\). Both divisibility conditions hold: \(12=4\cdot3=6\cdot2\).

2. However, \(ab=4\cdot6=24\), and 12 is not divisible by 24.

3. This single example shows that the situation asked about can happen.

Remember. Checking the two divisors separately does not always justify multiplying them.

Key idea. Begin with the resident whose existence is guaranteed.

1. We know that someone in the town has beautiful handwriting. Choose such a person.

2. This person cannot be a poet, because no poet has beautiful handwriting.

3. The person cannot be a boxer either: every boxer is a poet, and we have already ruled out being a poet.

4. So there is a resident who is neither a boxer nor a poet. That contradicts the visitor’s claim about every resident.

Remember. To disprove a claim about everybody, find one person who cannot fit it.

Key idea. Assume there is a hippo and follow the consequences.

1. The zoo is known to have a giraffe. Suppose it also has a hippo.

2. The rule about hippos and giraffes would then force it to have a rhino.

3. But a zoo with both a hippo and a rhino cannot have a giraffe. This contradicts the giraffe that is already there. So our assumption was impossible: there is no hippo.

4. Every zoo must have at least a rhino or a hippo. Since this zoo has no hippo, it must have a rhino.

5. A giraffe and a rhino, with no hippo, cause no conflict with any of the three rules.

Remember. A contradiction rules out the assumption that led to it.

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. Use a difference of squares on the two whole brackets.

1. The identity is \(U^2-V^2=(U-V)(U+V)\). Here take \(U=2a+b\) and \(V=a-2b\).

2. Find the difference carefully:

   \[U-V=(2a+b)-(a-2b)=2a+b-a+2b=a+3b.\]

3. Find the sum:

   \[U+V=(2a+b)+(a-2b)=3a-b.\]

4. Multiply the two expressions in factorised form. This gives \((a+3b)(3a-b)\).

Remember. A complicated-looking bracket can be treated as a single quantity. Be careful when subtracting a bracket.

Key idea. Express the required quantities using the known sum and product.

1. Expand the square of the sum:

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

   So \(x^2+y^2=7^2-2\cdot10=29\).

2. For the cubes, expand \((x+y)(x^2-xy+y^2)\). The mixed terms cancel, leaving \(x^3+y^3\).

3. Therefore

   \[x^3+y^3=(x+y)(x^2+y^2-xy)=7(29-10)=133.\]

Remember. Find an identity that uses what you know; solving for the individual numbers may be unnecessary.

Key idea. The expression is almost the expansion of a cube.

1. Expand:

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

2. Factor the two middle terms: \(3x^2y+3xy^2=3xy(x+y)\). Since \(x+y=1\), these terms together equal \(3xy\).

3. Thus \(x^3+y^3+3xy=(x+y)^3=1^3=1\).

Remember. Before substituting individual numbers, compare an expression with a familiar identity.

Key idea. Expand both sides and use the fact that real squares are non-negative.

1. The left side expands to

   \[2a^2+2b^2+2c^2+2ab+2bc+2ac.\]

2. The right side expands to

   \[a^2+b^2+c^2+2ab+2bc+2ac.\]

3. Cancel the same mixed terms on both sides, then subtract \(a^2+b^2+c^2\). The result is \(a^2+b^2+c^2=0\).

4. Each square is at least zero. If even one were positive, their sum would be positive. Therefore all three squares, and all three numbers, must be zero.

5. Substituting \(a=b=c=0\) makes both original sides zero, so this triple really works.

Remember. A sum of real squares is zero only when every square is zero.

Key idea. Equal sums and equal sums of squares force equal products.

1. Write the common sum as \(s=a+b=c+d\). Squaring gives

   \[s^2=a^2+b^2+2ab=c^2+d^2+2cd.\]

2. The sums of squares are equal by the question. Subtract them from the equality above to obtain \(2ab=2cd\), so \(ab=cd\). Call this common product \(p\).

3. From the cube identity,

   \[a^3+b^3=(a+b)^3-3ab(a+b)=s^3-3ps.\]

4. Exactly the same calculation gives \(c^3+d^3=s^3-3ps\). Hence the two sums of cubes are equal.

Remember. Two pairs with the same sum and product have the same sum of cubes.

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. Square the given sum before trying to find x itself.

1. Squaring the given equality gives

   \[\left(x+\frac1x\right)^2=9.\]

2. The middle term is \(2x\cdot(1/x)=2\), so

   \[x^2+2+\frac1{x^2}=9.\]

3. Subtract 2. Thus \(x^2+1/x^2=7\). We did not need the individual value of x.

Remember. A symmetric expression can often be found directly from another symmetric expression.

Key idea. Use the sum-of-cubes identity with reciprocal terms.

1. For any u and v, expanding \((u+v)^3\) gives

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

2. Here \(u=x\), \(v=1/x\). We know \(u+v=3\) and \(uv=1\).

3. Substitution gives

   \[x^3+\frac1{x^3}=3^3-3\cdot1\cdot3=27-9=18.\]

Remember. Knowing the sum and product of two quantities is enough to find their sum of cubes.

Key idea. Use the zero sum twice: first by squaring, then by substituting for c.

1. Square \(a+b+c=0\):

   \[a^2+b^2+c^2+2ab+2bc+2ca=0.\]

   Rearranging proves \(a^2+b^2+c^2=-2(ab+bc+ca)\).

2. For the cubes, write \(c=-(a+b)\). Then

   \[a^3+b^3+c^3=a^3+b^3-(a+b)^3.\]

3. Expand the last cube and cancel:

   \[a^3+b^3+c^3=-3a^2b-3ab^2=-3ab(a+b).\]

4. Since \(a+b=-c\), this equals \(3abc\), proving the second identity.

Remember. A relation between variables can make otherwise complicated expansions collapse.

Key idea. First find the mixed products, then square the known sum of squares.

1. By C1.3,

   \[ab+bc+ca=-\frac12(a^2+b^2+c^2)=-\frac12.\]

2. Square this relation:

   \[\begin{aligned}\frac14&=(ab+bc+ca)^2\\&=a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c).\end{aligned}\]

   The final term is zero, so \(a^2b^2+b^2c^2+c^2a^2=1/4\).

3. Now square \(a^2+b^2+c^2=1\):

   \[1=a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2).\]

4. The mixed part equals \(2\cdot(1/4)=1/2\). Subtracting it leaves \(a^4+b^4+c^4=1/2\).

Remember. When squaring creates mixed terms, look for a second identity that evaluates those terms.

Key idea. To prove expressions equal, factor their difference.

1. Use \(z=-(x+y)\) in the sum of squares:

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

   Hence \(x^2+xy+y^2=1\).

2. Subtract two of the expressions:

   \[\begin{aligned}(x^3-x)-(y^3-y)&=(x^3-y^3)-(x-y)\\&=(x-y)(x^2+xy+y^2-1)\\&=0.\end{aligned}\]

   Thus \(x^3-x=y^3-y\).

3. The hypotheses treat x, y and z in the same way. Repeating the argument with y and z gives \(y^3-y=z^3-z\). All three expressions are therefore equal.

Remember. Equality can be easier to prove by showing that a difference is zero.

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.