# SOLUTION: There exists an integer 'a' such that if a|2m+1 and/or a|(m^2+1) and/or a|(m+1)^2+1, then a|4n+7. Note: Anywhere from 1 - 3 of the assumptions can be used t

Algebra ->  Algebra  -> Proofs -> SOLUTION: There exists an integer 'a' such that if a|2m+1 and/or a|(m^2+1) and/or a|(m+1)^2+1, then a|4n+7. Note: Anywhere from 1 - 3 of the assumptions can be used t      Log On

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 Click here to see ALL problems on Proofs Question 516605: There exists an integer 'a' such that if a|2m+1 and/or a|(m^2+1) and/or a|(m+1)^2+1, then a|4n+7. Note: Anywhere from 1 - 3 of the assumptions can be used to prove 'a' divides 4n+7, so you can use a|2m+1 to prove a|4n+7, or you can use a|2m+1 and a|(m^2+1)to prove a|4n+7, or you can use a|2m+1, a|(m^2+1), and a|((m+1)^2+1) to prove a|4n+7, or any other combination. Answer by richard1234(5390)   (Show Source): You can put this solution on YOUR website!You used "m" and "n" everywhere without specifying what n is, so I will have to assume that you meant "m" everywhere... We want to show that there exists an integer "a" such that, for all m, if a divides any or all of those expressions, then a divides 4m+7. This problem appears trivial because we can let a = 1 and m be some integer, and we're done. Does a have to be greater than 1 (which you did not specify)?