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Perfect Square

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Number Man

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Nov 17, 2009, 4:00:02 PM11/17/09
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Let P(m) = (4m)^(4m-1) + 4m^2 + 1, where m is a positive integer.

Is it true that this expression is not a perfect square for any value
of m?

Ian Parker

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Nov 22, 2009, 2:30:02 PM11/22/09
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No it is not true. Let us take a large number r and let m=r^2

(4m)^(4m-1) = ((2r)^(4m-1))^2

Now let (2r)^(4m-1) = Q

If P(m) were a perfect square Sqrt(P(m))>Q

Now Next perfect square is (Q+1)^2 or Q^2 + 2Q + 1

Now if r,m are any size at all 2Q >> (4m)^2 (I presume you mean (4m)^2
an not 4m^2 as when we put m=1 we get 69 (not a perfect square. with
(4m)^2 we get 81 which is).

If 2Q >> (4m)^2 the proposition fails.


- Ian Parker

Number Man

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Nov 22, 2009, 8:30:02 PM11/22/09
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No, I mean 4*m^2 and not (4m)^2.

Thanks.

[Mod note: I think Ian interpreted the question as "find a value of m for which it isn't a perfect square", rather than the rather harder "show that it is never a square whatever m is" which is presumably what the OP meant.]

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