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Question 1: Prove that the difference of the squares of 2 consecutive odd numbers is always divisible by 8 |

Question 2: Prove the identity (n+3)^2 -(2n+21) ≡ (n+6)(n-2) |

Question 3: Prove that (8n+4)^2 -(8n-4)^2 is always a multiple of 4 |

Question 4: Prove that the sum of 4 consecutive numbers is always even |

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Question 1: (2n+3)^2-(2n+1)^2 =4n^2+12n+9-(4n^2+4n+1) =8n+8=8(n+1) therefore the difference of the squares of 2 consecutive odd numbers is always divisible by 8 |

Question 2: (n+3)^2 -(2n+21) =(n^2+6n +9)-(2n+21) =n^2+4n-12 =(n+6)(n-2) |

Question 3: (8n+4)^2 -(8n-4)^2 =(64n^2+64n +16)-(64n^2-64n +16) =128n=4 \times 32n Therefore (8n+4)^2 -(8n-4)^2 is a multiple of 4 |

Question 4: n+(n+1)+(n+2)+(n+3) =4n+6 =2(2n+3) Any multiple of 2 is always even, therefore the sum of 4 consecutive numbers is always even |