# 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.

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 is either an integer or is irrational.

### Proof

Suppose that = 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 is an integer.
We conclude that is either rational (in which case it's an integer), or is irrational.

## 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.)

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.

## Sum of two squares

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

`Hint  -  Solution`

`Hint  -  Answer  -  Solution`