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    I'm asked to prove that:

    For any natural numbers a, b, c with c greater than or equal to 2, there is a natural number n such that an^2 + bn + c is not a prime.

    I feel like my approach should be to let x be non-prime such that an^2 + bn + c = x. Then to show there is a solution n, in the natural numbers to an^2 + bn + (c - x) = 0, whilst somehow making use of the fact x can be wrote as the product of prime factors? Not sure where to get stuck in though..
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    Seems like quite a tricky gcse question
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    (Original post by marcus2001)
    I'm asked to prove that:

    For any natural numbers a, b, c with c greater than or equal to 2, there is a natural number n such that an^2 + bn + c is not a prime.

    I feel like my approach should be to let x be non-prime such that an^2 + bn + c = x. Then to show there is a solution n, in the natural numbers to an^2 + bn + (c - x) = 0, whilst somehow making use of the fact x can be wrote as the product of prime factors? Not sure where to get stuck in though..
    hey you can solve this by proof by contradiction. best thing to assume is that there are natural numbers , a, b ,c greater and equal to 2 , such that an^2 + bn + c is a prime.

    However if we take a=b=c=2 we have , 2n^2+2n+2=2(n^2+n+1) which is a contradiction ( as 2 being factor means it not a prime, remember a prime only has factor 1 and the number itself). hence we have proved your result

    NB:a for stuff like this do proof by contraction - assume the thing is true and give a counterexample. Hence the opposite must hold
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    Proof by contradiction would be a good way to do this, suppose that there does not exist a natural number n such that an^2 + bn + c is not a prime. Then, only 1 and an^2 + bn + c divides an^2 + bn + c

    However, since c \geq 2 the number an^2 + bn + c > 2 so it would be enough to show that there exists an n such that 2 divides an^2 + bn + c - which is a contradiction.
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    (Original post by Namige)
    Seems like quite a tricky gcse question
    Then why don't you help instead of uselessly posting that?
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    (Original post by Noble.)
    Proof by contradiction would be a good way to do this, suppose that there does not exist a natural number n such that an^2 + bn + c is not a prime. Then, only 1 and an^2 + bn + c divides an^2 + bn + c

    However, since c \geq 2 the number an^2 + bn + c > 2 so it would be enough to show that there exists an n such that 2 divides an^2 + bn + c - which is a contradiction.
    how would i find the n? i swear we use a.b.c for this?
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    ermm am i being a bit thick ? all you have to do is let n = c
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    (Original post by the bear)
    ermm am i being a bit thick ? all you have to do is let n = c
    LOL you're right! It should've been obvious from the fact c \geq 2 but typically I overcomplicated it.
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    (Original post by falcon pluse)
    how would i find the n? i swear we use a.b.c for this?
    Look at what 'the bear' posted, that's one way to answer this question in one line :lol:
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    Just proceed by noting that if we take n=c then,

    an^2 + bn + c = c(ac+b+1)

    which is clearly not prime since divisible by c.
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    (Original post by Rooinek)
    Just proceed by noting that if we take n=c then,

    an^2 + bn + c = c(ac+b+1)

    which is clearly not prime since divisible by c.
    Someone has already pointed this out, there's no need to bump a 9 hour old thread if you're just going to repeat what has already been said.
 
 
 
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