42 CHAPTER 2 STRATEGIES FOR INVESTIGATING PROBLEMS
has no positive integer solutions.Solution: We wish to show that the equality (4) cannot be true. So assume, to the
contrary, that (4) is true. If b^2 + b + 1 = a^2 , then b < a, and a^2 - b^2 = b + 1. As in
Example 2.2.7 on page 34, we employ the useful tactic of factoring, which yields
(a - b)(a+b) =b+l. (5)
Since a > b 2 1, we must have a - b 2 1 and a + b 2 2 + b, so the left-hand side of (5)
is greater than or equal to 1. (b + 2), which is strictly greater than the right-hand side
of (5). This is an impossibility, so the original assumption, that (4) is true, must in fact
be false. _Here is another impossibility proof, a classical argument from ancient Greece.
Example 2.3.2 Show that V2 is not rational.
Solution: Seeking a contradiction, let us suppose that V2 is rational. Then V2
can be expressed as a quotient of two positive integers (without loss of generality
we can assume both numerator and denominator are positive). Now we shall use the
extreme principle: of all the possible ways of doing this, pick the quotient for whichthe denominator is smallest.
Thus we write V2 = alb where a,b EN, where b is as small as possible. This
means the fraction alb is in "lowest terms," for if it were not, we could divide both a
and b by a positive integer greater than 1, making both a and b smaller, contradicting
the minimality of b. In particular, it is impossible that both a and b are even.
However, V2 = a I b implies that 2b^2 = a^2 , so a^2 is even (since it is equal to 2
times an integer). But this implies that a must be even as well (for if a were odd, a^2
would also be odd), and hence a is equal to 2 times an integer. So we can write a = 2t,
where tEN. Substituting, we get2 b^2 = a^2 = ( 2 t)^2 = 4t^2 ,
so b^2 = 2 t^2. But now by exactly the same reasoning, we conclude that b is also
even! This is impossible, so we have contradicted the original assumption, that V2
is rational. _Argument by contradiction can be used to prove "positive" statements as well.
Study the next example.Example 2.3.3 (Greece 19 95) If a,b, c,d,e are real numbers such that the equation
ax^2 + ( c+b)x+ (e+d) = 0
has real roots greater than 1, show that the equationax^4 +bx^3 +cx^2 +dx+e = 0
has at least one real root.