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.

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

Suppose that = r/s is rational.

Then r^{n} = ms^{n}.

By the Fundamental Theorem of Arithmetic, both sides of the equation must have the same prime factorization.

In the prime factorizations of r^{n} and s^{n}, 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 *n*th power, and is an integer.

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

Firstly, we note that if b = c = 0, then a = 0.

If c = 0 and b 0, then = −a/b, contradicting the above lemma. (1^{3} < 2 < 2^{3} 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

−2b^{3} | = a^{3} + 3a^{2}c + 3ac^{2}()^{2} + c^{3}()^{3} |

= a^{3} + 6ac^{2} + (3a^{2}c + 2c^{3}) |

Hence = −(2b^{3} + a^{3} + 6ac^{2})/(3a^{2}c + 2c^{3}). (Note that the denominator is necessarily non-zero.)

Again, this contradicts the lemma.

Therefore the only solution is 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} = x^{3} + y^{3} + 3xy(x + y), and cubing both sides, we obtain

−a^{3} | = 2b^{3} + 3c^{3} + 3bc(b + c) |

= 2b^{3} + 3c^{3} − 3abc |

Now = (a^{3} + 2b^{3} + 3c^{3})/3abc, contradicting the lemma.

Therefore the only solution is 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} = x^{3} + y^{3} + 3xy(x + y), and cubing both sides, we obtain

−a^{3} | = 2b^{3} + 4c^{3} + 6bc(b + c) |

= 2b^{3} + 4c^{3} − 6abc |

Hence a^{3} + 2b^{3} + 4c^{3} − 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 = 2a_{1}.

Then 8a_{1}^{3} + 2b^{3} + 4c^{3} − 12a_{1}bc = 0, from which 4a_{1}^{3} + b^{3} + 2c^{3} − 6a_{1}bc = 0.

Hence b is even. Let b = 2b_{1}.

Then 4a_{1}^{3} + 8b_{1}^{3} + 2c^{3} − 12a_{1}b_{1}c = 0, from which 2a_{1}^{3} + 4b_{1}^{3} + c^{3} − 6a_{1}b_{1}c = 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.

Show that a^{2} + b^{2} = 3c^{2} has no solution in positive integers.

Hint - Solution

Find a necessary and sufficient condition for one of the roots of x^{2} + ax + b = 0 to be the square of the other root.

Hint - Answer - Solution

- Infinite Descent
- Irrationality by Infinite Descent
- Fermat's Infinite Descent
- Infinite Descent versus Induction

Source: Traditional