Theorem 1. Two plus two equals four.

Proof of Theorem 1. The set of integers is an (infinite) group with respect to

addition. Since 2 is an integer, the sum of 2 and 2 must also be an integer.

Suppose, for the sake of contradiction, that 2 + 2 < 4. We have 2 > 0.

Adding two to both sides, we get 2+2 > 0+2. Since 0 is the identity element

for addition, we have 2 + 2 > 2. Hence

2 < 2 + 2 < 4

so 2 + 2 must equal 3. Since 3 is prime, by Fermat’s Little Theorem we

have the following for a ∈ Z

+:

a

3−1 ≡ 1 (mod 3)

a

3−1 ≡ 1 (mod 2 + 2)

But

a

3−1 =

a

3

a

=

a × a × a

a

= a × a

and, for a = 2,

2 × 2 ≡ 0 (mod 2 + 2)

since, by the definition of multiplication, 2 × 2 = 2 + 2. So, we have

2

3−1 ≡ 0 (mod 2 + 2)

2

2+2−1 ≡ 0 (mod 2 + 2)

This is a contradiction to Fermat’s Little Theorem, so 2 + 2 must not be

prime. But 3 is prime. Hence 2+2 must not equal 3, and therefore 2+2 ≥ 4.

It remains to show that 2 + 2 is not greater than 4. To prove this we need

the following lemma:

Lemma 1. ∀a ∈ Z, if a > 4, ∃b ∈ Z such that b > 0 and a − 2 = 2 + b.

If a were a solution to the equation 2 + 2 = a, then we would have

a − 2 = 2 + 0. The lemma states that this cannot hold for any a > 4, and so

a = 4, as desired. So, it only remains to prove our lemma.

Aaron Segal, 2010 2 + 2 = 4

Proof of Lemma 1. The proof is by induction over a. Our base case is a = 5.

Let 5−2 = 2+b. Five is the 5th Fibonacci number, and 2 is the 3rd Fibonacci

number. Therefore, by the definition of Fibonacci numbers, 5 − 2 must be

the 4th Fibonacci number. Letting fi denote the ith Fibonacci number, then

we have

fi − fi−1 > 0

for i 6= 2, because f2 − f1 = f0 and f0 = 0, but f1 = 1, and the Fibonacci

sequence is nondecreasing. Hence (5 − 2) − 2 > 0.

Now, suppose that ∃b ∈ Z such that b > 0 and (k − 1) − 2 = 2 + b. We

need to prove that, for some b

0 > 0, k − 2 = 2 + b

0

. Our inductive hypothesis

is equivalent to:

k − 1 − 2 = 2 + b

k − 1 − 2 + 1 = 2 + b + 1

k − 2 = 2 + (b + 1)

Since 1 > 0 and b > 0, we have (b + 1) > 0. Thus, letting b

0 = b + 1, we

have a nonnegative solution to k − 2 = 2 + b

0

, as desired.

By Lemma 1, it is not the case that 2 + 2 > 4. Hence 2 + 2 ≤ 4. We also

have 2 + 2 ≥ 4. Therefore,

2 + 2 = 4

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