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Special Factorizations: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

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Problem 1 of 10

Problem 1 Four factors and one small addition

Difficulty: 1 of 3 stars, Stretch

Prove that $(n+1)(n+3)(n+5)(n+7)+16$ is a perfect square for every integer $n$ . Give an explicit expression whose square it equals, including when $n$ is negative.

Builds on Difference of Squares, Factoring by Grouping
1. Hint 1
  Pair the first factor with the last, and the two middle factors with each other.
2. Hint 2
  The two resulting quadratic expressions differ by $8$ . Write them as $A-4$ and $A+4$ .
Answer

$(n+1)(n+3)(n+5)(n+7)+16=(n^2+8n+11)^2$ .

Full solution

Pairing the outside factors gives $(n+1)(n+7)=n^2+8n+7$ , while pairing the inside factors gives
$(n+3)(n+5)=n^2+8n+15.$
The linear term matches in both products.
Their average is $A=n^2+8n+11$ , so the two products are $A-4$ and $A+4$ .

The original expression therefore becomes
$(A-4)(A+4)+16=A^2-16+16=A^2.$
Substituting back proves the identity for every real $n$ , not just positive integers.

When $n$ is an integer, $A$ is also an integer, so its square is a perfect square.
A negative $n$ creates no exception: the identity still holds and the square of a negative integer is still a perfect square.
The choice of $16$ is explained by the grouping: it exactly cancels the square of the half-gap between the paired products.

Answer

$(n+1)(n+3)(n+5)(n+7)+16=(n^2+8n+11)^2$ .

Key idea

Look for pairs of factors with the same sum so that their products differ by a constant.
Problem 1: Mark complete
Problem 2 Two cube quotients

Difficulty: 1 of 3 stars, Stretch

Positive real numbers $a,b$ satisfy $a+b=7$ and $a\ne b$ . Find all ordered pairs $(a,b)$ for which
$\frac{a^3+b^3}{a+b}-\frac{a^3-b^3}{a-b}=-24.$
Prove that every pair you find satisfies the original equation.

Builds on Sum and Difference of Cubes, Sums and Products of Roots
1. Hint 1
  Factor the sum and difference of cubes before substituting any numbers.
2. Hint 2
  After cancellation, the difference of the two quotients is $-2ab$ . Combine the resulting product with $a+b=7$ .
Answer

$(a,b)=(3,4)$ or $(4,3)$ .

Full solution

The denominators are nonzero: positivity gives $a+b>0$ , and the problem states $a\ne b$ .
The cube factorizations yield
$\frac{a^3+b^3}{a+b}=a^2-ab+b^2,\qquad \frac{a^3-b^3}{a-b}=a^2+ab+b^2.$
Subtracting cancels the square terms and leaves $-2ab$ .
Thus the given equation is equivalent to $ab=12$ .

The sum and product now identify $a,b$ as roots of
$x^2-7x+12=(x-3)(x-4)=0.$
Since their sum is $7$ , the only ordered pairs are $(3,4)$ and $(4,3)$ .

Both pairs consist of unequal positive numbers and therefore lie in the original domain.
For either order, the first quotient is $9-12+16=13$ and the second is
$9+12+16=37.$
Their difference is $-24$ , as required.
Factoring first exposes the product information that the unsimplified cube expressions conceal.

Answer

$(a,b)=(3,4)$ or $(4,3)$ .

Key idea

Special factorizations can turn apparently cubic information into a simple sum-and-product reconstruction.
Problem 2: Mark complete
Problem 3 A denominator with three radicals

Difficulty: 1 of 3 stars, Stretch

Express
$\frac{1}{\sqrt2+\sqrt3+\sqrt6}$
in the form $A+B\sqrt2+C\sqrt3+D\sqrt6$ , where $A,B,C,D$ are rational. Give exact values and show your rationalization steps.

Builds on Rationalizing Denominators, Difference of Squares
1. Hint 1
  Treat $\sqrt2+\sqrt3$ as one block and $\sqrt6$ as another block.
2. Hint 2
  After the first conjugate, the denominator is $-1+2\sqrt6$ . A second conjugate will make it rational.
Answer

$A=-12/23$ , $B=7/23$ , $C=5/23$ , $D=-1/23$ .

Full solution

Group the denominator as $(\sqrt2+\sqrt3)+\sqrt6$ and multiply numerator and denominator by $(\sqrt2+\sqrt3)-\sqrt6$ .
The denominator becomes
$(\sqrt2+\sqrt3)^2-(\sqrt6)^2=(5+2\sqrt6)-6=-1+2\sqrt6.$
This quantity is positive, so the conjugate factor is not zero.

Now multiply by $1+2\sqrt6$ in both numerator and denominator.
The new denominator is
$(2\sqrt6-1)(2\sqrt6+1)=24-1=23.$
The numerator is
$(\sqrt2+\sqrt3-\sqrt6)(1+2\sqrt6).$

Expand and simplify the radical products: $2\sqrt{12}=4\sqrt3$ and
$2\sqrt{18}=6\sqrt2.$
The numerator is therefore $7\sqrt2+5\sqrt3-\sqrt6-12$ .
Division by $23$ gives the stated rational coefficients.
The two stages work because each conjugate eliminates one chosen sum, even though the first stage does not remove every radical.

Answer

$A=-12/23$ , $B=7/23$ , $C=5/23$ , $D=-1/23$ .

Key idea

With three radicals, rationalize in blocks; a useful first step need not finish the job.
Problem 3: Mark complete
Problem 4 Which symmetric factorization is possible?

Difficulty: 2 of 3 stars, Challenge

For a real number $k$ , seek real numbers $a,b$ such that the identity
$x^4+kx^2+16=(x^2+ax+b)(x^2-ax+b)$
holds for every real $x$ .

Find all possible $k$ , and for each such $k$ list every possible ordered pair $(a,b)$ . Include boundary cases where two listed formulas give the same pair.

Builds on Squares of Binomials, Difference of Squares, Factoring by Grouping
1. Hint 1
  View the right-hand side as a difference of squares before expanding.
2. Hint 2
  Compare constant terms and coefficients of $x^2$ . First split into the two possibilities for $b$ .
Answer

For every $k\le8$ : $(a,b)=(\pm\sqrt{8-k},4)$ . For $k\le-8$ there are also $(a,b)=(\pm\sqrt{-8-k},-4)$ . No pairs for $k>8$ ; a zero square root gives one pair, not two.

Full solution

The right-hand side is
$(x^2+b)^2-a^2x^2=x^4+(2b-a^2)x^2+b^2.$
Equality for every real $x$ requires $b^2=16$ and $2b-a^2=k$ .
These conditions are also sufficient, because they make the two expanded expressions identical.

If $b=4$ , then $a^2=8-k$ .
Real choices exist precisely when $k\le8$ , and they are
$a=\pm\sqrt{8-k}.$
If $b=-4$ , then $a^2=-8-k$ .
These choices exist precisely when $k\le-8$ , with
$a=\pm\sqrt{-8-k}.$
No other $b$ can satisfy the constant-term condition.

Thus $-8<k<8$ gives two pairs, $k=8$ gives the single pair $(0,4)$ , and $k<-8$ gives four pairs.
At $k=-8$ , the first branch gives $(4,4),(-4,4)$ and the second gives $(0,-4)$ , for three pairs.
There are no solutions for $k>8$ .
Keeping zero square roots as single choices avoids double-counting at the boundaries.

Answer

For every $k\le8$ : $(a,b)=(\pm\sqrt{8-k},4)$ . For $k\le-8$ there are also $(a,b)=(\pm\sqrt{-8-k},-4)$ . No pairs for $k>8$ ; a zero square root gives one pair, not two.

Key idea

An identity for every input becomes a coefficient system; its real solutions require explicit square-sign checks.
Problem 4: Mark complete
Problem 5 Composite but never a square

Difficulty: 2 of 3 stars, Challenge

For a positive integer $n$ , let $N=n^4+4n^2+16$ .

(a) Prove that $N$ is composite by finding two integer factors greater than $1$ .

(b) Prove that $N$ is never a perfect square.

Builds on Squares of Binomials, Difference of Squares
1. Hint 1
  For part (a), compare $N$ with $(n^2+4)^2$ and use a difference of squares.
2. Hint 2
  For part (b), compare $N$ with $(n^2+2)^2$ and the next integer square. Check the smallest value of $n$ separately.
Answer

$N=(n^2-2n+4)(n^2+2n+4)$ is composite and is never a perfect square.

Full solution

Add and subtract a suitable square:
$N=(n^2+4)^2-(2n)^2=(n^2-2n+4)(n^2+2n+4).$
The smaller factor is $(n-1)^2+3\ge3$ , and the larger is
$(n+1)^2+3\ge7.$
Both are integers greater than $1$ , proving that $N$ is composite.

For square values, use a different nearby square:
$N=(n^2+2)^2+12.$
When $n\ge2$ , the gap to the next integer square is
$(n^2+3)^2-(n^2+2)^2=2n^2+5\ge13.$
Therefore
$(n^2+2)^2<N<(n^2+3)^2.$
There is no perfect square strictly between consecutive integer squares.

The remaining case is $n=1$ , which gives $N=21$ , strictly between $4^2=16$ and $5^2=25$ .
Thus no positive integer $n$ gives a square.
Different square comparisons serve different purposes: one produces factors, while the other excludes square values.

Answer

$N=(n^2-2n+4)(n^2+2n+4)$ is composite and is never a perfect square.

Key idea

Choose a nearby square to fit the goal: factorization and square exclusion may need different choices.
Problem 5: Mark complete
Problem 6 Recover the two radicands

Difficulty: 2 of 3 stars, Challenge

Find all ordered pairs of positive integers $(a,b)$ satisfying
$\sqrt a+\sqrt b=\sqrt{24+8\sqrt5}.$
Prove that your list is complete; do not merely guess a way to simplify the right-hand radical. All square roots are nonnegative real square roots.

Builds on Squares of Binomials, Fractional Exponents and Radicals, Sums and Products of Roots
1. Hint 1
  Square once and use the integers $S=a+b$ and $P=ab$ to describe the result.
2. Hint 2
  Isolate $2\sqrt P$ and square again. If $S\ne24$ , the resulting equation would express $\sqrt5$ as a rational number.
Answer

$(a,b)=(4,20)$ or $(20,4)$ .

Full solution

Let $S=a+b$ and $P=ab$ , both positive integers.
Squaring the original equation gives $S+2\sqrt P=24+8\sqrt5$ , hence
$2\sqrt P=24-S+8\sqrt5.$
Squaring again yields
$4P=(24-S)^2+320+16(24-S)\sqrt5.$

If $S\ne24$ , this equation would express $\sqrt5$ as a ratio of integers, contradicting its irrationality.
To recall why $\sqrt5$ is irrational, a reduced positive fraction $u/v$ with square $5$ would give $u^2=5v^2$ .
The prime $5$ would divide $u$ ; substituting $u=5w$ would then show it also divides $v$ , a contradiction.
Thus $S=24$ .

The first squared equation now gives $\sqrt P=4\sqrt5$ , so $P=80$ .
Therefore $a,b$ are the roots of
$x^2-24x+80=(x-4)(x-20)=0.$
Their sum forces the two ordered pairs $(4,20)$ and $(20,4)$ .

For either pair, the left side is $2+2\sqrt5$ , a positive number whose square is $24+8\sqrt5$ .
It therefore equals the specified nonnegative square root on the right.
This final sign check verifies the original equation after the two squarings.

Answer

$(a,b)=(4,20)$ or $(20,4)$ .

Key idea

When matching a radical to a sum of square roots, justify the rational and irrational parts separately and verify the final sign.
Problem 6: Mark complete
Problem 7 Do not divide away a family

Difficulty: 2 of 3 stars, Challenge

Find all ordered pairs of real numbers $(a,b)$ satisfying
$a^3+b^3=7(a+b),\qquad a^2+b^2=10.$
Prove that your list is complete.

Builds on Sum and Difference of Cubes, Sums and Products of Roots
1. Hint 1
  Factor $a^3+b^3$ and use the second equation to simplify the quadratic factor.
2. Hint 2
  You should reach $(a+b)(3-ab)=0$ . Treat $a+b=0$ separately from $ab=3$ .
Answer

$(\sqrt5,-\sqrt5)$ , $(-\sqrt5,\sqrt5)$ , $(1,3)$ , $(3,1)$ , $(-1,-3)$ , and $(-3,-1)$ .

Full solution

The sum-of-cubes identity and $a^2+b^2=10$ transform the first equation into
$(a+b)(10-ab)=7(a+b).$
Equivalently,
$(a+b)(3-ab)=0.$
We must consider both factors; division by $a+b$ could remove valid solutions.

If $a+b=0$ , then $b=-a$ .
The second equation becomes $2a^2=10$ , giving
$a=\pm\sqrt5.$
This yields the first two ordered pairs.

If $ab=3$ , then $(a+b)^2=a^2+b^2+2ab=16$ , so $a+b=4$ or $a+b=-4$ .
For sum $4$ , the numbers are roots of $x^2-4x+3=(x-1)(x-3)$ , yielding $(1,3),(3,1)$ .
For sum $-4$ , the equation is $x^2+4x+3=(x+1)(x+3)$ , yielding $(-1,-3),(-3,-1)$ .

All six pairs have square sum $10$ .
In the first branch their cube sum and their sum are both zero.
In the second, the identity gives
$a^3+b^3=(a+b)(10-3)=7(a+b).$
Thus every pair works, and the exhaustive factor split proves completeness.

Answer

$(\sqrt5,-\sqrt5)$ , $(-\sqrt5,\sqrt5)$ , $(1,3)$ , $(3,1)$ , $(-1,-3)$ , and $(-3,-1)$ .

Key idea

A common factor may describe an entire solution family; split into cases before canceling it.
Problem 7: Mark complete
Problem 8 Two cubes, one difference

Difficulty: 3 of 3 stars, Deep challenge

Find all pairs of positive integers $(a,b)$ with $a>b$ such that $a^3-b^3=728$ . Prove that no pair is missing.

Builds on Sum and Difference of Cubes, Factoring Quadratics
1. Hint 1
  Let $d=a-b$ . Express the difference of cubes in terms of $b$ and $d$ , and obtain an upper bound for $d$ .
2. Hint 2
  The difference is even, so $a$ and $b$ have the same parity. Use the even divisors of $728$ below $9$ , then solve or eliminate each case.
Answer

$(a,b)=(12,10)$ or $(9,1)$ .

Full solution

Let $d=a-b>0$ .
Factoring gives
$728=d(a^2+ab+b^2)=d(3b^2+3bd+d^2).$
Since $b>0$ , the right-hand side exceeds $d^3$ .
Thus $d^3<728<9^3$ , so $d\le8$ .
Also, an integer cube has the same parity as its base.
The even difference $728$ means $a,b$ have the same parity, and therefore $d$ is even.

Now $d$ is an even positive divisor of $728=8\cdot7\cdot13$ and is at most $8$ .
The only possibilities are $2,4,8$ .
If $d=2$ , the equation reduces to $b^2+2b-120=(b+12)(b-10)=0$ , so positivity gives $b=10,a=12$ .

If $d=4$ , division by $4$ gives $182=3b^2+12b+16$ , or
$166=3b^2+12b.$
The right-hand side is divisible by $3$ and the left is not, so this case is impossible.
If $d=8$ , the equation becomes $b^2+8b-9=(b+9)(b-1)=0$ , giving $b=1,a=9$ .

Finally, $12^3-10^3=1728-1000=728$ and
$9^3-1^3=729-1=728.$
The bound and divisor list make the search complete.

Answer

$(a,b)=(12,10)$ or $(9,1)$ .

Key idea

In an integer equation, factor first, bound a small difference, and only then enumerate cases.
Problem 8: Mark complete
Problem 9 When a cubic expression has a minimum

Difficulty: 3 of 3 stars, Deep challenge

Fix a real number $S$ . Real numbers $a,b,c$ may vary subject to $a+b+c=S$ . Consider
$E=a^3+b^3+c^3-3abc.$
For each of the cases $S>0$ , $S=0$ , and $S<0$ , determine whether $E$ has a least possible value. If it does, find that value and all equality cases. If it does not, prove that $E$ can be made arbitrarily negative.

Builds on Sum and Difference of Cubes, Factoring by Grouping, Squares of Binomials
1. Hint 1
  Try factoring $E$ by $a+b+c$ . The remaining quadratic expression can be written using the three squares $(a-b)^2,(b-c)^2,(c-a)^2$ .
2. Hint 2
  For negative $S$ , try $a=S/3+t$ , $b=S/3-t$ , $c=S/3$ , where $t$ can be as large as needed.
Answer

If $S>0$ , minimum $0$ only at $a=b=c=S/3$ . If $S=0$ , every permitted triple gives $E=0$ . If $S<0$ , there is no minimum; $E$ is unbounded below.

Full solution

Expanding and canceling the mixed cubic terms verifies
$E=(a+b+c)(a^2+b^2+c^2-ab-bc-ca).$
The second factor equals half the sum of the three squared differences.
Under the sum constraint, therefore,
$E=\frac S2\big((a-b)^2+(b-c)^2+(c-a)^2\big).$

If $S>0$ , every square is nonnegative, so $E\ge0$ .
Equality holds exactly when all three differences vanish.
The constraint then forces $a=b=c=S/3$ , which does give $E=0$ .
If $S=0$ , the factored expression is zero for every allowed triple; equality of the variables is not necessary.

If $S<0$ , use the family $a=S/3+t$ , $b=S/3-t$ , $c=S/3$ .
Its sum is always $S$ , and the squared differences add to $6t^2$ .
Consequently $E=3St^2$ .
The coefficient $3S$ is negative, so taking positive $t$ sufficiently large makes this value less than any specified negative number.
Thus there is no least value in this case.
The sign of the fixed sum controls the entire conclusion.

Answer

If $S>0$ , minimum $0$ only at $a=b=c=S/3$ . If $S=0$ , every permitted triple gives $E=0$ . If $S<0$ , there is no minimum; $E$ is unbounded below.

Key idea

A cubic expression can become a signed sum of squares after the right factorization.
Problem 9: Mark complete
Problem 10 A parameter that vanishes

Difficulty: 3 of 3 stars, Deep challenge

Let $a,b,c$ be pairwise distinct real numbers, and let $t$ be any real number. Evaluate
$\frac{(a-t)^2}{(a-b)(a-c)}+\frac{(b-t)^2}{(b-a)(b-c)}+\frac{(c-t)^2}{(c-a)(c-b)}.$
Your answer must hold for every permitted choice of the four numbers. Prove it by algebra.

Builds on Factoring by Grouping, Difference of Squares, Algebraic Fractions
1. Hint 1
  Use the common denominator $(a-b)(a-c)(b-c)$ . Watch the signs when reversing a difference.
2. Hint 2
  After expanding in $t$ , its squared and linear terms cancel. Group the remaining numerator first by $b-c$ .
Answer

The expression always equals $1$ .

Full solution

Pairwise distinctness makes every denominator nonzero.
Let
$D=(a-b)(a-c)(b-c).$
Rewriting all three fractions over $D$ gives the numerator
$N=(a-t)^2(b-c)-(b-t)^2(a-c)+(c-t)^2(a-b).$
The minus sign in the middle term comes from
$b-a=-(a-b).$

Expand only the squares first.
The coefficient of $t^2$ is
$(b-c)-(a-c)+(a-b)=0.$
The coefficient of $-2t$ is
$a(b-c)-b(a-c)+c(a-b)=0.$
Hence $N$ is independent of $t$ and equals $a^2(b-c)-b^2(a-c)+c^2(a-b)$ .

Group this remaining expression by powers of $a$ :
$N=a^2(b-c)-a(b^2-c^2)+bc(b-c).$
Factoring $b^2-c^2=(b-c)(b+c)$ now yields
$N=(b-c)\big(a^2-a(b+c)+bc\big)=(b-c)(a-b)(a-c)=D.$

Thus the original sum is $N/D=1$ .
Every division used a nonzero difference, and no restriction on $t$ was needed.
The calculation explains both why the parameter disappears and why the remaining value is constant.

Answer

The expression always equals $1$ .

Key idea

When several fractions have related difference factors, a common denominator can expose systematic cancellation.
Problem 10: Mark complete