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Question 1: Prove that the difference of the squares of 2 consecutive even numbers is always divisible by 4 |

Question 2: Prove that (3n+4)^2 +(3n-4)^2+1 is always odd |

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

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

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Question 1: (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 2: (3n+4)^2 +(3n-4)^2+1 =(9n^2+24n +16)+(9n^2-24n +16)+1 =18n^2+32+1=2(9n^2+16)+1 (3n+4)^2 +(3n-4)^2 is a multiple of 2, therefore it is even. Adding 1 to an even number will always make it odd. |

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

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 |