Solution to puzzle 103: Root sums

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

• a + b + c = 0.
• a + b + c = 0.
• a + b + c = 0.

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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.

i.  a + b + c = 0

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

If c 0, we rewrite the equation as −b = a + c.  Cubing both sides, we obtain

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

Hence = −(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 + b + c = 0

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

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

−a3	= 2b3 + 3c3 + 3bc(b + c)
	= 2b3 + 3c3 − 3abc

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

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

iii.  a + b + c = 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 = b + c.  Making use of the identity (x + y)³ = x³ + y³ + 3xy(x + y), and cubing both sides, we obtain

−a3	= 2b3 + 4c3 + 6bc(b + c)
	= 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.

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Sum of two squares

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

Hint  -  Solution

Quadratic roots

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

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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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