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[Magic Math]

Prove that for $n\in N$,
$\sqrt{n+3}+\sqrt{n}$ is rational if and only if $n=1$.

I arrive at proving that $n^2+3n$ cannot be the square of an integer for $n>1$.
That is, there does not exist an interger $k$ such that
$$ n^2+3n = k^2 $$
How to prove this?

[nooonuii]

ข้อความเดิมเขียนโดยคุณ Magic Math

Prove that for $n\in N$,
$\sqrt{n+3}+\sqrt{n}$ is rational if and only if $n=1$.

Suppose that $\sqrt{n+3}+\sqrt{n}$ is rational. Then

$\dfrac{3}{\sqrt{n+3}+\sqrt{n}}=\sqrt{n+3}-\sqrt{n}$ is also rational.

Thus $\sqrt{n+3} = \dfrac{(\sqrt{n+3}+\sqrt{n})+(\sqrt{n+3}-\sqrt{n})}{2}$ is rational. Similarly $\sqrt{n}$ is also rational.

Let $\sqrt{n+3}=r,\sqrt{n}=s$. Argue that $r,s$ are nonnegative integers. Then we have

$(r-s)(r+s)=3$ which implies

$r-s = \pm 1,\pm 3$
$r+s=\pm 3,\pm 1$.

Then we must have that $2r=4,-4$ which implies $r=2$ and hence $n=1$.

[Magic Math]

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[Magic Math]

Why r and s are integers? The square root of a positive integer need not be an integer.
I think that $r-s$ and $r+s$ are nonnegative rationals, NOT necessarily integers.
For $a,b \in Q^+$, $a \cdot b =3$ does not always imply $a=3,-3$ or $a=-1,1$. For example, $a=1/2,b=6$.

[nooonuii]

ข้อความเดิมเขียนโดยคุณ Magic Math

Why r and s are integers? The square root of a positive integer need not be an integer.

It is just a fact from elementary number theory.

Suppose $r=\dfrac{a}{b}$. Then

$(n+3)b^2=a^2$ implies $b^2|a^2$.

Thus $b|a$ which means $r$ is an integer.

The same idea for $s$.

[Magic Math]

Thank you very much for your discussion.

[kongp]

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