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Question 1: Prove that the difference of the squares of 2 consecutive numbers is always the sum of the 2 numbers |

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

Question 3: Prove the identity (n+2)^2 -(3n+16) ≡ (n+4)(n-3) |

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

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Question 1: (2n+1)^2-(2n)^2 =4n^2+4n+1-4n^2 =4n+1=(2n+1)+(2n) therefore the difference of the squares of 2 consecutive numbers is always the sum of the 2 numbers |

Question 2: (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 3: (n+2)^2 -(3n+16) =(n^2+4n +4)-(3n+16) =n^2+n-12 =(n+4)(n-3) |

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 |