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Question 1: Prove that (9n+3)^2 +(9n-3)^2+1 is always odd |

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

Question 3: Prove the identity (n+3)^2 -(3n+27) ≡ (n+6)(n-3) |

Question 4: Prove that the product of 2 consecutive odd numbers is always odd |

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Question 1: (9n+3)^2 +(9n-3)^2+1 =(81n^2+54n +9)+(81n^2-54n +9)+1 =162n^2+18+1=2(81n^2+9)+1 (9n+3)^2 +(9n-3)^2 is a multiple of 2, therefore it is even. Adding 1 to an even number will always make it odd. |

Question 2: (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 3: (n+3)^2 -(3n+27) =(n^2+6n +9)-(3n+27) =n^2+3n-18 =(n+6)(n-3) |

Question 4: (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 |