# 13, sec. 4.5/4.6
For any integer n, n2 – 2 is not divisible by 4.
The negation: There exists integer n such that n2 – 2 is divisible by 4.
Proof:
Suppose not. That is, there exists integer n such that n2 – 2 is divisible by 4. By the definition of
divisibility there is some integer m such that
𝑛2 – 2 = 4𝑚
Adding 2 to both sides of the equation, and using addition and distributive properties of equalities we
obtain
(1)
𝑛2 = 4𝑚 + 2 = 2(2𝑚 + 1).
where 2𝑚 + 1 is an integer. Therefore, 𝑛2 is even by the definition of an even number.
According to Proposition 4.5.4 (4.6.4 full edition) of sec. 4.5 (4.6 full edition) 𝑛 is also even. Thus
𝑛 = 2𝑘
for some integer 𝑘 (by the definition of even numbers). Squaring both sides of the latter equation we get
(2)
𝑛2 = 4 𝑘 2 .
Applying the transitive property of equalities to equations (1) and (2) we have
2(2𝑚 + 1) = 4 𝑘 2 .
Dividing both sides of the equation by 2 we obtain
2𝑚 + 1 = 2 𝑘 2
Here 𝑘 2 and 𝑚 are integers. Using the definitions of odd and even numbers, we obtain that odd number
2𝑚 + 1 equals even number 2𝑘 2. But the result contradicts the Parity Property according to which every
integer is either even or odd. Therefore for any integer n, n2 – 2 is not divisible by 4.
#14, sec. 4.5/4.6
For all prime numbers, a, b, c, 𝒂𝟐 + 𝒃𝟐 ≠ 𝒄𝟐
The negation: there exist prime numbers, a, b, c, 𝒂𝟐 + 𝒃𝟐 = 𝒄𝟐
Proof:
Suppose not. That is, there exist prime numbers 𝑎, 𝑏, 𝑐, such that 𝑎2 + 𝑏 2 = 𝑐 2 .
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According to the definition of a prime number a, b, and c are integers greater than 1.
Subtracting b2 from both sides of the previous equation we obtain
(1)
𝑎2 = 𝑐 2 − 𝑏 2 = (𝑐 − 𝑏)(𝑐 + 𝑏)
Since 𝑎2 and 𝑐 + 𝑏 are positive integers, 𝑐 − 𝑏 must be greater than zero. Otherwise, 𝑎2 ≠ (𝑐 − 𝑏)(𝑐 +
𝑏).
Let’s assume 𝑐 − 𝑏 > 0.
There are three possible factorizations of 𝑎2 :
𝑎2 = 𝑎 ∙ 𝑎 = 𝑎2 ∙ 1 = 1 ∙ 𝑎2
Case I: Let 𝑎2 = 𝑎 ∙ 𝑎 . Using the uniqueness of prime factorization and equation (1) we obtain
𝑐 − 𝑏 = 𝑎, 𝑐 + 𝑏 = 𝑎.
Solving the system of equations we find that 𝑐 = 𝑎 and 𝑏 = 0 which contradicts the supposition that b is
prime.
Case II: Let 𝑎2 = 𝑎2 ∙ 1. From equation (1), we obtain that −𝑏 = 𝑎2 , 𝑐 + 𝑏 = 1 . The latter equation
cannot be satisfied, because c and b are both greater than 0.
Case III: Let 𝑎2 = 1 ∙ 𝑎2 . Using equation (1) we obtain 𝑐 + 𝑏 = 𝑎2 , 𝑐 − 𝑏 = 1. If 𝑐 − 𝑏 = 1, then c and
b are consecutive integers. Due to the parity property, one of the numbers is even and the other one is
odd. There is only one even prime number 2 with the consecutive prime integer 3. But if c=3 and b=2
and c+b=5. Since 5 is not a perfect square, a is not an integer which contradicts the assumption that a is
prime.
Therefore, For all prime numbers, a, b, c, 𝒂𝟐 + 𝒃𝟐 ≠ 𝒄𝟐
#15, sec. 4.5
If a, b, c are integers and 𝑎2 + 𝑏 2 = 𝑐 2 , then at least one of a and b is even.
The negation: there are integers a, b, c such that 𝑎2 + 𝑏 2 = 𝑐 2 and a and b are both odd.
Proof:
Suppose not. That is, there are integers a, b, c such that 𝑎2 + 𝑏 2 = 𝑐 2 and a and b are both odd.
Using the definition of odd integers for 𝑎 and 𝑏 we obtain
𝑎 = 2𝑛 + 1 𝑎𝑛𝑑 𝑏 = 2𝑚 + 1
where 𝑛and 𝑚 are integers.
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Making the substitutions and using the rules of algebra we have
𝑐 2 = 𝑎2 + 𝑏 2 = (2𝑛 + 1)2 + (2𝑚 + 1)2 = 2(2𝑛2 + 2𝑛 + 2𝑚2 + 2𝑚 + 1)
Since 2𝑛2 + 2𝑛 + 2𝑚2 + 2𝑚 + 1 is an integer as a sum and product of integers, 𝑐 2 is even by the
definition of an even number.
According to Proposition 4.5.4 (or 4.6.4), if c2 is even, then c is even. Applying the definition of even
numbers to 𝑐 we have
𝑐 = 2𝑘.
where 𝑘 is integer. Squaring both sides we receive
𝑐 2 = 4𝑘 2
Since 𝑐 2 = 2(2𝑛2 + 2𝑛 + 2𝑚2 + 2𝑚 + 1) and 𝑐 2 = 4𝑘 2 , using the transitive property of equality we
obtain
2(2𝑛2 + 2𝑛 + 2𝑚2 + 2𝑚 + 1) = 4𝑘 2
Dividing both sides of the equation by 2 we have
(1)
2𝑛2 + 2𝑛 + 2𝑚2 + 2𝑚 + 1 = 2𝑘 2
Let 𝑠 = 𝑛2 + 𝑛 + 𝑚2 + 𝑚. Then equation (1) can be written as
2𝑠 + 1 = 2𝑘 2
Here 𝑠 and 𝑘 2 are integers (as sum or product of integers). Using the definition of odd and even numbers
we obtain that an odd integer 2𝑠 + 1 equals even integer 2𝑘 2 which contradicts the Parity Property.
Therefore, if a, b, c are integers and 𝒂𝟐 + 𝒃𝟐 = 𝒄𝟐 , then at least one of a and b is even.
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