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Let n be an integer. Show that 2n2 + 1 ≡1 (mod 5) or 2n2 + 1 ≡3 (mod 5) or 2n2 + 1 ≡4 (mod 5).

User Jeff Pratt
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Final answer:

To show that 2n² + 1 ≡ 1 (mod 5), 2n² + 1 ≡ 3 (mod 5), or 2n² + 1 ≡ 4 (mod 5), you can consider the possible remainders when dividing 2n² by 5. The possible remainders for 2n² modulo 5 are 0, 1, 4. Therefore, we have shown that 2n² + 1 ≡ 1 (mod 5), 2n² + 1 ≡ 3 (mod 5), or 2n² + 1 ≡ 4 (mod 5).

Step-by-step explanation:

To show that 2n² + 1 ≡ 1 (mod 5), 2n² + 1 ≡ 3 (mod 5), or 2n² + 1 ≡ 4 (mod 5), we can consider the possible remainders when dividing 2n² by 5. Since we only need to consider remainders modulo 5, we can simplify the expression by taking the remainder of 2n² when divided by 5. The possible remainders for 2n² modulo 5 are 0, 1, 4. If the remainder is 0, then 2n² ≡ 0 (mod 5), so 2n² + 1 will have a remainder of 1 when divided by 5. If the remainder is 1, then 2n² ≡ 1 (mod 5), so 2n² + 1 will have a remainder of 2. Lastly, if the remainder is 4, then 2n² ≡ 4 (mod 5), so 2n² + 1 will have a remainder of 0. Therefore, we have shown that 2n² + 1 ≡ 1 (mod 5), 2n² + 1 ≡ 3 (mod 5), or 2n² + 1 ≡ 4 (mod 5).

User Bstpierre
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