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If $d\mid2n^2$, could $n^2+d$ be a square of a natural number? Here $d, n \in \Bbb N$.

What I tried so far: $2n^2 = kd, k \in N$ $$d = 2\frac{n^2}{k}$$ $$n^2 + d = $$ $$n^2 +2\frac{n^2}{k}= n^2\left(\frac{k+2}{k}\right) = \frac{kd}{2}\left(\frac{k+2}{k}\right) = \frac{d(k+2)}{2}$$

I don't see where this could lead me.

G Tony Jacobs
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Shocky2
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2 Answers2

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Your reasoning looks good. You have that $$n^2+d=n^2\left(\frac{k+2}{k}\right)$$

That is a perfect square only if $\frac{k+2}{k}$ is the square of a rational number. However, a rational square, in lowest terms, is a ratio of two integer squares. No two integers squares differ by $2$, and a possible reduction of $\frac{k+2}{k}$ wouldn't help, because no two integer squares differ by $1$. (Obviously, $0$ and $1$ aren't helpful here.)

We conclude that $n^2+d$ is not the square of an integer.

G Tony Jacobs
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  • Are you saying that $n^2q$ is perfect square of natural number iff $q$ is perfect square of rational number? – nonuser Nov 20 '17 at 19:11
  • Yes, indeed. We have $n^2q=m^2$ if and only if $q=\left(\frac{m}{n}\right)^2$ – G Tony Jacobs Nov 20 '17 at 19:17
  • I mean..... assuming $n\ne 0$. That would be an exception. Not sure if we're counting $0$ among the natural numbers. – G Tony Jacobs Nov 20 '17 at 19:17
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    And assuming that $n^2q$ is an integer (which is given here because it's equal to $n^2+d$). – Daniel Fischer Nov 20 '17 at 19:18
  • Right...... I see. The implication goes one way. If the integer $n^2q$ is the square of an integer $m$, then we have that $q$ is the square of a rational number. Did I say it wrong in my answer? Yes, I did. Now I've made an edit to fix that. Thank you. – G Tony Jacobs Nov 20 '17 at 19:22
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I am assuming, for this answer, that $0\not\in\mathbb{N}$.

The Reasoning in the Question

The question asks whether there are $m,n,d\in\mathbb{N}$ so that $d\mid2n^2$, and $n^2+d=m^2$. This is equivalent to asking if there are $m,n\in\mathbb{N}$ so that $m\gt n$ and $m^2-n^2\mid2n^2$.

We can assume that $(m,n)=1$. If not, then divide $m$ and $n$ by $(m,n)$.

Suppose that $$ m^2-n^2\mid2n^2\tag1 $$ This means there is a $k$ so that $$ k\left(m^2-n^2\right)=2n^2\tag2 $$ which means that $$ (k+2)n^2=km^2\tag3 $$ Finishing with Bezout

Since $(m,n)=1$, this answer says we also have $\left(m^2,n^2\right)=1$. Bezout gives us $a,b$ so that $am^2+bn^2=1$. Therefore, $$ \begin{align} k &=akm^2+bkn^2\tag{4a}\\ &=a(k+2)n^2+bkn^2\tag{4b}\\ &=(ak+2a+bk)n^2\tag{4c} \end{align} $$ Explanation:
$\text{(4a):}$ multiply $am^2+bn^2=1$ by $k$
$\text{(4b):}$ apply $(3)$
$\text{(4c):}$ collect multiples of $n^2$

$(4)$ says that $n^2\mid k$. Then, dividing $(2)$ by $n^2$ gives $$ \frac{k}{n^2}\left(m^2-n^2\right)=2\tag5 $$ That is, $m^2-n^2\mid2$. This means that either $m^2-n^2=2$ or $m^2-n^2=1$.
If $2=m^2-n^2=(m+n)(m-n)$, we must have $m+n=2$ and $m-n=1$ which means $m=\frac32$ and $n=\frac12$.
If $1=m^2-n^2=(m+n)(m-n)$, we must have $m+n=1$ and $m-n=1$ which means $m=1$ and $n=0$.

Since neither $\frac12$ nor $0$ are in $\mathbb{N}$, we cannot have $m,n\in\mathbb{N}$ and $m^2-n^2\mid2n^2$.

A Comment on the Other Answer

While the claim made in G Tony Jacobs' answer, that if $\frac ab=\left(\frac pq\right)^2$ and $(a,b)=1$ then $a$ and $b$ are squares of integers, is true, I think it requires some justification.

Suppose that $(a,b)=1$ and $(p,q)=1$ and $$ \frac ab=\left(\frac pq\right)^2\tag6 $$ Bezout gives us $x,y$ such that $$ ax+by=1\tag7 $$ We have $$ \begin{align} p^2 &=ap^2x+bp^2y\tag{8a}\\ &=ap^2x+aq^2y\tag{8b}\\ &=a\left(p^2x+q^2y\right)\tag{8c} \end{align} $$ Explanation:
$\text{(8a):}$ multiply $(7)$ by $p^2$
$\text{(8b):}$ apply $(6)$
$\text{(8c):}$ factor out $a$

Similarly, we have $$ \begin{align} q^2 &=aq^2x+bq^2y\tag{9a}\\ &=bp^2x+bq^2y\tag{9b}\\ &=b\left(p^2x+q^2y\right)\tag{9c} \end{align} $$ As mentioned above $(p,q)=1\implies\left(p^2,q^2\right)=1$. Thus, since $(8)$ and $(9)$ say that $p^2x+q^2y$ is a common factor of $p^2$ and $q^2$, we must have $$ p^2x+q^2y=1\tag{10} $$ Then $(8)$, $(9)$, and $(10)$ say that $a=p^2$ and $b=q^2$.

robjohn
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