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Solution to puzzle 103: Root sums

Skip restatement of puzzle.Let a, b, c be rational numbers.  Show that each of the following equations can be satisfied only if a = b = c = 0.


First of all, note that we may assume a, b, c are integers, for we can multiply each equation by the least common multiple of the denominators.

We will proceed by reductio ad absurdum; assume that each equation can be satisfied for non-zero a, b, c, and derive a contradiction.  For this we will need the following lemma.

Lemma

Let n, m be positive integers, with n > 1.  Then nth root of m is either an integer or is irrational.

Proof

Suppose that nth root of m = r/s is rational.
Then rn = msn.
By the Fundamental Theorem of Arithmetic, both sides of the equation must have the same prime factorization.
In the prime factorizations of rn and sn, each prime occurs a multiple of n times.
Hence each prime in the prime factorization of m must occur a multiple of n times.
That is, m is an nth power, and nth root of m is an integer.
We conclude that nth root of m is either rational (in which case it's an integer), or is irrational.

i.  a + bcube root of 2 + croot 2 = 0

Firstly, we note that if b = c = 0, then a = 0.
If c = 0 and b not equal to 0, then cube root of 2 = −a/b, contradicting the above lemma.  (13 < 2 < 23 implies cube root of 2 is not an integer, and so must be irrational.)

If c not equal to 0, we rewrite the equation as −bcube root of 2 = a + croot 2.  Cubing both sides, we obtain

−2b3 = a3 + 3a2croot 2 + 3ac2(root 2)2 + c3(root 2)3
  = a3 + 6ac2 + (3a2c + 2c3)root 2

Hence root 2 = −(2b3 + a3 + 6ac2)/(3a2c + 2c3).  (Note that the denominator is necessarily non-zero.)
Again, this contradicts the lemma.

Therefore the only solution is a = b = c = 0.

ii.  a + bcube root of 2 + ccube root of 3 = 0

Note that if a = 0, then b = c = 0, for otherwise cube root of 3/2 is rational, and so 2 × cube root of 3/2 = cube root of 12 is rational, contradicting the above lemma.
If b = 0, then a = c = 0, for otherwise cube root of 3 = −a/c is rational.
If c = 0, then a = b = 0, for otherwise cube root of 2 = −a/b is rational.

Now suppose a, b, c are non-zero, and rewrite the equation as −a = bcube root of 2 + ccube root of 3.  Making use of the identity (x + y)3 = x3 + y3 + 3xy(x + y), and cubing both sides, we obtain

−a3 = 2b3 + 3c3 + 3bccube root of 6(bcube root of 2 + ccube root of 3)
  = 2b3 + 3c3 − 3abccube root of 6

Now cube root of 6 = (a3 + 2b3 + 3c3)/3abc, contradicting the lemma.

Therefore the only solution is a = b = c = 0.

iii.  a + bcube root of 2 + ccube root of 4 = 0

As above, we can conclude that if one of a, b, c equals zero, then all must equal zero.  We now assume a, b, c are all non-zero.

Rewrite the equation as −a = bcube root of 2 + ccube root of 4.  Making use of the identity (x + y)3 = x3 + y3 + 3xy(x + y), and cubing both sides, we obtain

−a3 = 2b3 + 4c3 + 6bc(bcube root of 2 + ccube root of 4)
  = 2b3 + 4c3 − 6abc

Hence a3 + 2b3 + 4c3 − 6abc = 0.(1)

We now assume that the greatest common divisor of a, b, c is 1.  (If not, we can divide a, b, c by their greatest common divisor.)

From (1), a is even.  Let a = 2a1.
Then 8a13 + 2b3 + 4c3 − 12a1bc = 0, from which 4a13 + b3 + 2c3 − 6a1bc = 0.
Hence b is even.  Let b = 2b1.
Then 4a13 + 8b13 + 2c3 − 12a1b1c = 0, from which 2a13 + 4b13 + c3 − 6a1b1c = 0.
We now conclude that c is even, so that a, b, c are all even.  This is a contradiction, as we assumed that the greatest common divisor of a, b, c is 1.

Therefore the only solution is a = b = c = 0.


Sum of two squares (1 star)

Show that a2 + b2 = 3c2 has no solution in positive integers.

Hint  -  Solution

Quadratic roots (2 star)

Find a necessary and sufficient condition for one of the roots of x2 + ax + b = 0 to be the square of the other root.

Hint  -  Answer  -  Solution

Further reading

  1. Infinite Descent
  2. Irrationality by Infinite Descent
  3. Fermat's Infinite Descent
  4. Infinite Descent versus Induction

Source: Traditional

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