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**Is this proof rigorous enough?**

## Homework Statement

This problem is from my College Problem Solving Seminar Course.

Book:

__The Art and Craft of Problem Solving__, 2nd Edition by Paul Zeitz.

Ex. 1.2.1

Prove that the product of four consecutive natural numbers cannot be the square of an integer.

## Homework Equations

Number Theory

## The Attempt at a Solution

What I need help with is just critique of my proof. I want to make sure it is very rigorous (and ofcourse correct).

Attempt at a Solution:

Let,

[tex]

{n \in \mathbb{N}}

[/tex]

[tex]

{\mathbb{N} = \{1, 2, 3, . . . ,\infty\}

[/tex]

[tex]

{m \in \mathbb{Z}}

[/tex]

[tex]

{\mathbb{Z} = \{ - \infty, . . . , - 1, 0, 1, . . . , \infty\}

[/tex]

Prove that:

[tex]

{n(n + 1)(n + 2)(n + 3)} \neq {{m}^{2}}

[/tex]

(Proof by Contradiction)

Assume the Contrary:

[tex]

{n(n + 1)(n + 2)(n + 3)} = {{m}^{2}}

[/tex]

[tex]

{{n}^{4} + 6{n}^{3} + 11{n}^{2} + 6{n}} = {{m}^{2}}

[/tex]

Let,

[tex]

f(n) = {{n}^{4} + 6{n}^{3} + 11{n}^{2} + 6{n}}

[/tex]

Then, since [tex]f(n)[/tex] is a fourth degree polynomial. Then for,

[tex]

{{n}^{4} + 6{n}^{3} + 11{n}^{2} + 6{n}} = {{m}^{2}}

[/tex]

there must exist some [tex]g(n)[/tex] such that,

[tex]

{f(n)} = {[g(n)]}^{2}

[/tex]

If [tex]g(n)[/tex] exists, then it must be a second degree polynomial--since [tex]f(n)[/tex] is a fourth degree polynomial--fitting one of the following forms:

[tex]

{(1) g(n) = a{n}^{2}}

[/tex]

[tex]

{(2) g(n) = a{n}^{2}} + c

[/tex]

[tex]

{(3) g(n) = a{n}^{2}} + b{n}

[/tex]

[tex]

{(4) g(n) = a{n}^{2}} + b{n} + c

[/tex]

Consider case (1), this is not possible for [tex]g(n)[/tex] since [tex]f(n)[/tex] would lack a term of degree three.

Consider case (2), this is not possible for [tex]g(n)[/tex] since [tex]f(n)[/tex] would also lack a term of degree three.

Consider case (3), this is not possible for [tex]g(n)[/tex] since [tex]f(n)[/tex] would lack a term of degree one.

Consider case (4), this is the only possibility for [tex]g(n)[/tex] since [tex]f(n)[/tex] would contain terms of degrees: four, three, two, and one.

However, in order for,

[tex]

g(n) = a{n}^{2}} + b{n} + c

[/tex]

It must satisfy,

[tex]

f(n) = {(a{n}^{2}} + b{n} + c)}^{2}

[/tex]

Which when expanded is,

[tex]

f(n) = {{a^2}{n}^{4} + {2ab}{n}^{3} + {(2ac + b^2)}{n}^{2} + {2bc}{n} + {c^2}}

[/tex]

Where,

[tex]

f(n) = {{n}^{4} + 6{n}^{3} + 11{n}^{2} + 6{n} + 0}

[/tex]

Must match as follows,

[tex]

a^2 = 1

[/tex]

[tex]

2ab = 6

[/tex]

[tex]

2ac + b^2 = 11

[/tex]

[tex]

2bc = 6

[/tex]

[tex]

c^2 = 0

[/tex]

However if,

[tex]

c^2 = 0

[/tex]

Then,

[tex]

2bc \neq 6

[/tex]

Contradiction.

However, here is my question...

Is there anything I missed in this proof? Anything I should have done more rigorously?

Anything I left unaddressed?

Thanks,

-PFStudent

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