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I know that n!+1 is a perfect square. But the equation below leads to that as well.

((n-1)*(n+1))+1 = n^2 is a perfect square where n=integer.

But is also works for fractions.
for example, the square root of 17 = 4.12310..., and (3.12310... * 5.12310...)+1 = 17

Is there a proof for this?

Similarly, there is a formula for cubes that seems to work for all number, and creates a perfect cube for integers:

((n-1)n(n+1))+n = n^3 if n=integer is a perfect cube.
but works for all values of n

proof??

Pavook
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    if $n=3$ then $3!+1=7$ is not a perfect square...what do you mean? – lulu Dec 02 '16 at 17:58
  • Do you know that $n!+1$ is a perfect square? Note that $3!+1=7.$ On the other hand, use that $(n-1)(n+1)=n^2-1.$ So $(n-1)(n+1)+1=n^2.$ Also $(n-1)n(n+1)=n^3-n.$ Thus $(n-1)n(n+1)+n=n^3.$ – mfl Dec 02 '16 at 17:59
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    But is also works for fractions $(x-1)(x+1)+1=x^2$ is an algebraic identity, and it works for all numbers. You could let $x = \pi$ and it would work as well. – dxiv Dec 02 '16 at 18:10
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    @dxiv More precisely those polynomial equations remain true when evaluated in any ring. Generally a polynomial equation remains true when evaluated in any ring where the variable values and coefficients all commute. This is the universal property of polynomial rings, e.g. $, x^2 - y^2 = (x-y)(x+y),$ remains true when evaluated in any ring where $,x,y,$ commute. So it is not generally true in noncommutative rings, e.g. when $,x,y,$ are matrices, or difference or differential operators, etc. – Bill Dubuque Dec 02 '16 at 18:29

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As already pointed out, $n!+1$ is not necessarily a perfect square .. in fact it rarely is (it happens to be true for $n=5$, but that is one of the rare cases).

But you are correct to say that $(n-1)(n+1)+1$ is always a perfect square. The proof is simple: for any $n$: $(n-1)(n+1)+1 = n^2 + n - n -1 + 1 = n^2$

And for the last formula you get: $(n-1)n(n+1)+n = n((n-1)(n+1)+1) =$ (as we just saw) $ = nn^2 = n^3$

Bram28
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