[size=150]Proposition If n ∈ N, then 1[sup]2[/sup] | (n[sup]4[/sup] − n[sup]2[/sup]).[br][br]Proof. We will prove [i]l[/i]this with strong induction.[br][br](1) First note that the statement is true for the first six positive integers:[br]For n = 1, 1[sup]2[/sup] divides 1[sup]4[/sup] −1[sup]2[/sup] = 0.[br]For n = 2, 1[sup]2[/sup] divides 2[sup]4[/sup] −2[sup]2[/sup] = 12.[br]For n = 3, 1[sup]2[/sup] divides 3[sup]2[/sup] −3[sup]2[/sup] = 72.[br]For n = 4, 1[sup]2[/sup] divides 4[sup]4[/sup] −4[sup]2[/sup] = 240.[br]For n = 5, 1[sup]2[/sup] divides 5[sup]4[/sup] −5[sup]2[/sup] = 600.[br]For n = 6, 1[sup]2[/sup] divides 6[sup]4[/sup] −6[sup]2[/sup] = 1260.[br][br](2) For k ≥ 6, assume 12 | (m[sup]4[/sup] − m[sup]2[/sup]) for 1 ≤ m ≤ k (i.e., S[sub]1[/sub], S[sub]2 [/sub],..., S[sub]k[/sub] are true).[br]We must show S[sub]k+1[/sub] is true, that is, 12 |(k +1)[sup]4[/sup] −(k +1)[sup]2[/sup].[br]Now, Sk−5being true means 12 |(k −5)[sup]4[/sup] −(k −5)[sup]2[/sup]. To simplify, put k −5 = [i]l[/i] so[br]12 | ([i]l[sup]4 [/sup][/i][i]- l[/i][sup]2[/sup] )meaning ([i]l[/i][sup]4[/sup][i] - l[/i][sup]2[/sup] )=12a for a ∈ Z, and k +1 = [i]l[/i] +6 . Then:[br][br](k +1)[sup]4[/sup] −(k +1)[sup]2[/sup] = ([i]l[/i]+6)[sup]4[/sup] −([i]l[/i]+6)[sup]2[/sup][br]= [i]l[sup]4[/sup][/i] +24[i]l[sup]3[/sup][/i] +216[i]l[sup]2[/sup][/i] +864[i]l[/i]+1296−([i]l[sup]2[/sup][/i] +12[i]l[/i]+36)[br]= ([i]l[sup]4[/sup][/i] −[i]l[sup]3[/sup][/i])+24[i]l[sup]3[/sup][/i] +216[i]l[sup]2[/sup][/i] +852[i]l[/i]+1260[br]= 12a+24[i]l[sup]3[/sup][/i] +216[i]l[sup]2[/sup][/i] +852[i]l[/i]+1260[br]= 12¡a+2[i]l[/i]3 +18[i]l[/i]2 +71[i]l[/i]+105.[br][br]Because (a+2[i]l[sup]3[/sup][/i] +18[i]l[sup]2[/sup][/i] +71[i]l[/i]+105) ∈ Z, we get 12 |(k+1)[sup]4[/sup]−(k+1)[sup]2[/sup].[/size]

[size=150]To prove the proposition "If n ∈ N, then 1[sup]2[/sup] | (n[sup]4[/sup] − n[sup]2[/sup])" using strong induction, the proof is presented as follows:[br][br](1) Base case: Show that the statement is true for the first six positive integers.[br]The base case demonstrates that the proposition holds for small values of n.[br][br]For n = 1, we have 1[sup]2[/sup] divides 1[sup]4[/sup] − 1[sup]2[/sup] = 0.[br]For n = 2, we have 1[sup]2[/sup] divides 2[sup]4[/sup] − 2[sup]2[/sup] = 12.[br]For n = 3, we have 1[sup]2[/sup] divides 3[sup]4[/sup] − 3[sup]2[/sup] = 72.[br]For n = 4, we have 1[sup]2[/sup] divides 4[sup]4[/sup] − 4[sup]2[/sup] = 240.[br]For n = 5, we have 1[sup]2[/sup] divides 5[sup]4[/sup] − 5[sup]2[/sup] = 600.[br]For n = 6, we have 1[sup]2[/sup] divides 6[sup]4[/sup] − 6[sup]2[/sup] = 1260.[br][br](2) Inductive step: Assume the proposition is true for all positive integers from 1 to k, where k ≥ 6.[br]In the inductive step, the assumption is that the proposition holds for all positive integers from 1 to k. We will show that the proposition is true for k + 1.[br][br]To prove the proposition for k + 1, consider the following steps:[br][br]Step 1: Use the assumption from the inductive hypothesis.[br]Assume that 1[sup]2[/sup] | (m[sup]4[/sup] − m[sup]2[/sup]) for 1 ≤ m ≤ k, which means S[sub]1[/sub], S[sub]2[/sub], ..., S[sub]k[/sub] are true.[br][br]Step 2: Express (k + 1)[sup]4[/sup] − (k + 1)[sup]2[/sup] in terms of the inductive hypothesis.[br]Now, consider the expression (k + 1)[sup]4[/sup] − (k + 1)[sup]2[/sup]. We can rewrite k + 1 as (k − 5) + 6, where 6 is chosen because the base case showed the proposition holds for k = 6. So, we set k − 5 = l, giving us k + 1 = l + 6.[br][br]Step 3: Simplify the expression using the inductive hypothesis.[br]Now, rewrite the expression as follows:[br](k + 1)[sup]4[/sup] − (k + 1)[sup]2[/sup] = (l + 6)[sup]4[/sup] − (l + 6)[sup]2[/sup][br]= [i]l[sup]4[/sup][/i] + 24[i]l[sup]3[/sup][/i] + 216[i]l[sup]2[/sup][/i] + 864l + 1296 − ([i]l[sup]2[/sup][/i] + 12l + 36)[br]= ([i]l[sup]4[/sup][/i]− [i]l[sup]3[/sup][/i]) + 24[i]l[sup]3[/sup][/i] + 216[i]l[sup]2[/sup][/i] + 852[i]l[/i] + 1260[br][br]Step 4: Factor out 12 from the expression.[br]We observe that [i]l[sup]4[/sup][/i] −[i]l[sup]3[/sup][/i] is divisible by 12 due to the inductive hypothesis. Let ([i]l[sup]4[/sup][/i] − [i]l[sup]3[/sup][/i]) = 12a for some a ∈ Z.[br][br]So, the expression becomes:[br](k + 1)[sup]4[/sup] − (k + 1)[sup]2[/sup] = 12a + 24[i]l[sup]4[/sup][/i] + 216[i]l[sup]3[/sup][/i] + 852[i]l[/i] + 1260[br][br]Step 5: Further simplification.[br]We can factor out 12 from the remaining terms, which will leave us with a multiple of 12:[br](k + 1)[sup]4[/sup] − (k + 1)[sup]2[/sup] = 12(a + 2[i]l[sup]4[/sup][/i] + 18[i]l[sup]3[/sup][/i] + 71[i]l[/i] + 105)[br][br]Since (a + 2l[sup]3[/sup] + 18l[sup]2[/sup] + 71[i]l[/i] + 105) ∈ Z, we conclude that 1[sup]2[/sup] | (k + 1)[sup]4[/sup] − (k + 1)[sup]2[/sup].[br][br]This completes the inductive step, showing that the proposition is true for all positive integers. Thus, we have proven that if n ∈ N, then 1[sup]2[/sup] | (n[sup]4[/sup] − n[sup]2[/sup]).[/size]