> On 17 Nov, 21:00, Number Man <ab
...@balamand.edu.lb> wrote:
> > 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?
> 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
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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