[size=150]Proposition If n ∈ N, then 4 | (5[sup]n[/sup] −1).[br][br]Proof. We use proof by smallest counterexample. (We will number the steps[br]to match the outline, but that is not usually done in practice.)[br][br](1) If n = 1, then the statement is 4 | (5[sup]1[/sup] −1), or 4 | 4, which is true.[br](2) For sake of contradiction, suppose it’s not true that 4 | (5[sup]n[/sup] −1) for all n.[br](3) Let k > 1 be the smallest integer for which 4 - (5[sup]k[/sup] −1).[br](4) Then 4 | (5[sup]k[/sup]−1 −1), so there is an integer a for which 5[sup]k[/sup]−1 −1 = 4a. Then[br][br]5[sup]k[/sup]−1 −1 = 4a[br]5(5[sup]k[/sup]−1 −1) = 5·4a[br]5[sup]k[/sup] −5 = 20a[br]5[sup]k[/sup]−1 = 20a+4[br]5[sup]k[/sup] −1 = 4(5a+1).[br][br]This means 4 | (5[sup]k[/sup]−1), a contradiction, because 4 - (5[sup]k[/sup]−1) in Step 3. Thus,[br]we were wrong in Step 2 to assume that it is untrue that 4 | (5[sup]n[/sup] −1) for[br]every n. Therefore 4 | (5[sup]n[/sup] −1) is true for every n.[/size]

[size=150]The given proof is a proof by contradiction using the method of smallest counterexample to show that for every natural number n, 4 | (5[sup]n[/sup] - 1). Let's break down the steps of the proof:[br][br]1. Base case: Check the statement for n = 1.[br]The base case shows that the proposition holds true for at least one natural number.[br][br]For n = 1, we have 4 | (5[sup]1[/sup] - 1) = 4 | 4, which is true.[br][br]2. Assumption for contradiction: Assume the proposition is false for some n.[br]The proof begins by assuming that there exists a counterexample, i.e., there is at least one natural number n for which 4 does not divide (5[sup]n[/sup] - 1).[br][br]3. Smallest counterexample: Find the smallest integer k for which the proposition fails.[br]Among all the natural numbers where the proposition is false, let k be the smallest such number.[br][br]4. Express k in terms of a new integer a.[br]Since the proposition is false for k, we have 4 does not divide (5[sup]k[/sup] - 1). Thus, there is an integer a for which 5[sup]k[/sup] - 1 = 4a.[br][br]5. Rewrite the expression for k in terms of a.[br]Now, manipulate the equation 5[sup]k[/sup] - 1 = 4a to express k in terms of a:[br]5[sup]k[/sup] - 1 = 4a[br]5(5[sup]k[/sup] - 1) = 5 * 4a[br]5[sup]k[/sup] - 5 = 20a[br]5[sup]k[/sup] - 1 = 20a + 4[br]5[sup]k[/sup] - 1 = 4(5a + 1).[br][br]6. Reach a contradiction.[br]Since 4 divides (5[sup]k[/sup] - 1) (as expressed in step 5), it contradicts the assumption made in step 3 that k is the smallest counterexample.[br][br]7. Conclude the proof.[br]The contradiction shows that our assumption in step 2 was wrong, and there is no smallest counterexample. Therefore, the proposition "4 | (5[sup]n[/sup] - 1)" is true for every natural number n.[br][br]In conclusion, the proof establishes that for every natural number n, 4 divides (5[sup]n[/sup] - 1).[/size]