# Download Theorem: Let x and y be integers. is even if and only if Proof

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```Theorem: Let x and y be integers. (𝑥 + 1) ∗ 𝑦 2 is even if and only if 𝑥 is odd or 𝑦 is even.
Proof: Assume that x and y are integers. First we will prove that if (𝑥 + 1) ∗ 𝑦 2 is even, then
𝑥 is odd or 𝑦 is even. We will prove this by proving its contrapositive. The contrapositive
states that if x is even and y is odd, then (𝑥 + 1) ∗ 𝑦 2 is odd. So, assume that x is even and y
is odd. This means there exist integers m and n such that,
𝑥 = 2𝑚 and
𝑦 = 2𝑛 + 1.
By substitution,
(𝑥 + 1)𝑦 2 = (2𝑚 + 1)(2𝑛 + 1)2
= (2𝑚 + 1)(4𝑛2 + 4𝑛 + 1)
= 8𝑚𝑛2 + 8𝑚𝑛 + 2𝑚 + 4𝑛2 + 4𝑛 + 1
= 2(4𝑚𝑛2 + 4𝑚𝑛 + 𝑚 + 2𝑛2 + 2𝑛) + 1.
Note that (4𝑚𝑛2 + 4𝑚𝑛 + 𝑚 + 2𝑛2 + 2𝑛) is an integer and thus, (𝑥 + 1)𝑦 2 is an odd
integer. Therefore, if (𝑥 + 1)𝑦 2 is even, then x is odd or y is even. Now we will prove that if
x is odd or y is even, then (𝑥 + 1)𝑦 2 is even. We will prove this using cases. For the first
case, assume that x is odd. This means there exists some integer m such that
𝑥 = 2𝑚 + 1.
By substitution,
(𝑥 + 1)𝑦 2 = (2𝑚 + 1 + 1)𝑦 2
= 2𝑚𝑦 2 + 2𝑦 2
= 2(𝑚𝑦 2 + 𝑦 2 ).
Note that 𝑚𝑦 2 + 𝑦 2 is an integer and thus, (𝑥 + 1)𝑦 2 is even. For the next case we will
assume that y is even. This means there exists some integer n such that,
𝑦 = 2𝑛.
By substitution,
(𝑥 + 1)𝑦 2 = (𝑥 + 1)(2𝑛)2
= (𝑥 + 1)(4𝑛2 )
= 4𝑥𝑛2 + 4𝑛2
= 2(2𝑥𝑛2 + 2𝑛2 ).
Note that 2𝑥𝑛2 + 2𝑛2 is an integer and thus (𝑥 + 1)𝑦 2 is even. Therefore, (𝑥 + 1)𝑦 2 is even
if and only if x is odd or y is even, for all integers x and y. QED.
Proposition: For all integers a and b, (a + b)3 a3 + b3 (mod 3).
Proof: Let a and b be integers. There exists an integer g such that
(𝑎3 + 𝑏 3 ) + 3𝑔 = (𝑎 + 𝑏)3 , by the definition of cogruence
= (𝑎2 + 2𝑎𝑏 + 𝑏 2 )(𝑎 + 𝑏)
= 𝑎3 + 𝑏 3 + 2𝑎2 𝑏 + 2𝑎𝑏 2 + 𝑎𝑏 2 + 𝑎2 𝑏.
By simplifying the other side of the equation we get
3𝑔 = (𝑎 + 𝑏)3 − (𝑎3 + 𝑏 3 )
= 2𝑎2 𝑏 + 2𝑎𝑏 2 + 𝑎𝑏 2 + 𝑎2 𝑏
= 𝑎2 (2𝑏 + 𝑏) + 𝑏 2 (2𝑎 + 𝑎)
= 𝑎2 𝑏(2 + 1) + 𝑏 2 𝑎(2 + 1)
= 3(𝑎2 𝑏) + 3(𝑏 2 𝑎).
Let 𝑠 = 𝑎2 𝑏 and 𝑦 = 𝑏 2 𝑎, and note that s and y are integers by closure. Using substitution we obtain,
3𝑔 = 3𝑠 + 3𝑦.
We know that any number multiplied by itself must be divisible by itself, so the equation above follows
(a + b)3 a3 + b3 (mod 3). Thus, we have proven that for all integers a and b, (a + b)3 a3 + b3 (mod 3).



Theorem. For all real numbers x and y, |𝑥 − 𝑦| = |𝑦 − 𝑥|.
Proof: Suppose x and y are real numbers. Let 𝑎 = 𝑥 − 𝑦. We will prove by cases.
Case 1: Suppose 𝑎 ≥ 0. By the definition of absolute value, |𝑎| = 𝑎, so
|𝑥 − 𝑦| = 𝑥 − 𝑦. Also, |−𝑎| = −(−𝑎) = 𝑎, so |𝑦 − 𝑥| = 𝑥 − 𝑦. Thus |𝑎| = |−𝑎| and
|𝑥 − 𝑦| = |𝑦 − 𝑥|.
Case 2: Suppose 𝑎 < 0. By the definition of absolute value, |𝑎| = −𝑎, so
|𝑥 − 𝑦| = 𝑦 − 𝑥. Also, |−𝑎| = −𝑎 since 𝑎 < 0, so |𝑦 − 𝑥| = 𝑦 − 𝑥. Thus |𝑎| = |−𝑎| and
|𝑥 − 𝑦| = |𝑦 − 𝑥|.
So |𝑥 − 𝑦| = |𝑦 − 𝑥| for all real numbers x and y. 
Prove: Let a and b be integers with 𝑎 ≠ 0. If a does not divide b, then the equation 𝑎𝑥 3 + 𝑏𝑥 +
(𝑏 + 𝑎) = 0 does not have a solution that is a natural number.
Proof: Let a and b be integers with 𝑎 ≠ 0. We will show this by proving the contrapositive.
Contrapositive: If the equation 𝑎𝑥 3 + 𝑏𝑥 + (𝑏 + 𝑎) = 0 has a natural number solution,
then a divides b.
Let x be a natural number solution of 𝑎𝑥 3 + 𝑏𝑥 + (𝑏 + 𝑎) = 0. By rearranging the equation, we
obtain
𝑎𝑥 3 + 𝑏𝑥 + (𝑏 + 𝑎) = 0
𝑏𝑥 + 𝑏 = −𝑎𝑥 3 − 𝑎
𝑏(𝑥 + 1) = 𝑎(−𝑥 3 − 1)
−𝑥 3 −1
𝑏 = 𝑎(
𝑥+1
).
Since 𝑛 ∈ ℕ, 𝑥 + 1 ≠ 0 so we can divide both sides by 𝑥 + 1 to obtain
𝑏 = 𝑎(−𝑥 2 + 𝑥 − 1).
Note that (−𝑥 2 + 𝑥 − 1) ∈ ℤ by closure. Thus, a divides b by the definition of divides. So, we
have shown that the contrapositive is true.
Thus, if a does not divide b, then the equation 𝑎𝑥 3 + 𝑏𝑥 + (𝑏 + 𝑎) = 0 does not have a solution
that is a natural number. ■
```
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