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Question 1: Prove the identity (n+3)^2 -(5n+39) ≡ (n+6)(n-5) |

Question 2: Prove that the product of 2 consecutive even numbers is always a multiple of 4 |

Question 3: Prove that the difference of the squares of 2 consecutive even numbers is always divisible by 4 |

Question 4: Prove that (9n+8)^2 -(9n-8)^2 is always a multiple of 8 |

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Question 1: (n+3)^2 -(5n+39) =(n^2+6n +9)-(5n+39) =n^2+n-30 =(n+6)(n-5) |

Question 2: 2n(2n+2) =4n^2+4n =4(n^2+n) Therefore the product of 2 consecutive even numbers is always a multiple of 4 |

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

Question 4: (9n+8)^2 -(9n-8)^2 =(81n^2+144n +64)-(81n^2-144n +64) =288n=8 \times 36n Therefore (9n+8)^2 -(9n-8)^2 is a multiple of 8 |