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Question 1: Prove that (4n+6)^2 -(4n-6)^2 is always a multiple of 4 |

Question 2: Prove that the product of 2 consecutive odd numbers is always odd |

Question 3: Prove the identity (n+4)^2 -(4n+48) ≡ (n+8)(n-4) |

Question 4: Prove that the difference of the squares of 2 consecutive numbers is always the sum of the 2 numbers |

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Question 1: (4n+6)^2 -(4n-6)^2 =(16n^2+48n +36)-(16n^2-48n +36) =96n=4 \times 24n Therefore (4n+6)^2 -(4n-6)^2 is a multiple of 4 |

Question 2: (2n+1)(2n+3) =4n^2+8n+3 =2(2n^2+4n)+3 Any multiple of 2 is always even therefore 2(2n^2+4n) is even. Even+Odd=Odd therefore the product of 2 consecutive odd numbers is always odd |

Question 3: (n+4)^2 -(4n+48) =(n^2+8n +16)-(4n+48) =n^2+4n-32 =(n+8)(n-4) |

Question 4: (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 |