# Prove that $\sqrt{5n+2}$ is irrational

I'm trying to follow this answer to prove that $\sqrt{5n+2}$ is irrational. So far I understand that the whole proof relies on being able to prove that $(5n+2)|x^2 \implies (5n+2)|x$ (which is why $\sqrt{4}$ doesn't fit, but $\sqrt{7}$ etc. does), this is where I got stuck. Maybe I'm overcomplicating it, so if you have a simpler approach, I'd like to know about it. :)

A related problem I'm trying to wrap my head around is: Prove that $\frac{5n+7}{3n+4}$ is irreducible, i.e. $(5n+7)\wedge(3n+4) = 1$.

• what is $n$ here? – Dr. Sonnhard Graubner Dec 5 '16 at 19:30
• What is $n\wedge m$? – Dietrich Burde Dec 5 '16 at 19:32
• @DietrichBurde it's $\gcd(n,m)$ (and $n\lor m$ would be $\operatorname{lcm}(n,m)$) – user384138 Dec 5 '16 at 19:39
• Beware that the accepted answer is incorrect (or incomplete). If you wish to learn how to generalize the linked proof to your example then one way is to use the method in my answer. – Bill Dubuque Dec 5 '16 at 21:16

Well, one way to say it is irrational is to see that $5n+2$ isn't an integer square for any $n\in\mathbb{Z}$ (it only finish in $2$ or $7$). The other way you're trying lets you the same ending $(p,q\in \mathbb{Z}, q\neq 0)$: \begin{align*} \sqrt{5n+2}=\frac{p}{q}&& \\ 5n+2=\frac{p^2}{q^2} &&(1)\\ q^2(5n+2)=p^2 && (2) \end{align*}

Let $$p=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_t^{\alpha_t}$$ $$q=q_1^{\beta_1}q_2^{\beta_2}\dots q_s^{\beta_s}$$

where $p_i,q_j$ are primes and $\alpha_i,\beta_j$ are positive integers.

From $(1)$, $p^2/q^2$ is an integer (is equal to $5n+2$), so you have that the $q_i$ are certain primes $p_j$. Renaming the prime factors in a way such that $q_i=p_i$, you can let you the fraction (considering that $t>s$)

$$\frac{p^2}{q^2}=\frac{p_1^{2\alpha_1}p_2^{2\alpha_2}\dots p_t^{2\alpha_t}}{q_1^{2\beta_1}q_2^{2\beta_2}\dots q_s^{2\beta_s}}=p_1^{2(\alpha_1-\beta_1)}p_2^{2(\alpha_2-\beta_2)}\dots p_t^{2(\alpha_s-\beta_s)}\dots p_t^{2\alpha_t}=5n+2$$

then this implies that $5n+2$ is a square, but by the original statement, it can't be.

• Can anyone tell me why was downvoted my answer? – MonsieurGalois Dec 5 '16 at 19:56
• I feel like the "This implies that $5n+2$ is a square" part is the cause for downvotes. This is far from obvious for a person unfamiliar with number theory. – Wojowu Dec 5 '16 at 21:06
• Perhaps because your proof that $\,\sqrt{5n+2}\not\in \Bbb Q\,$ is either incomplete or incorrect. It's impossible to tell which is the case from what little you wrote. Nor does your answer tell how to do it the way the OP asks, using the linked proof. I explain both in my answer. – Bill Dubuque Dec 5 '16 at 21:06
• Please tell me if this is correct now. – MonsieurGalois Dec 5 '16 at 21:42
• @MonsieurGalois an easier way to see that 5n+2 is not an integer square is by seeing $(\frac{2}{5}) = (-1)^{\frac{5^2-1}{8}} = -1$, so 2 is not a quadratic residue modulo 5. – bat_of_doom Dec 26 '16 at 10:58

For the second question: $\frac{5n+7}{3n+4}$ is irreducible.

For that is enough to see that if there is a prime $p$ such that

$p|5n+7$ and $p|3n+4$

then

$p|[3(5n+7)-5(3n+4)] \Rightarrow p|1$ what is impossible.

• I explain a linear algebra genesis of the above in my answer. – Bill Dubuque Dec 5 '16 at 20:38

Hint $\ {\rm mod}\,\ 5\!:\,\ k^2 \in \{0,\pm1,\pm2\}^2\equiv \{0, \pm 1\},\,$ none $\equiv 2,\$ i.e. squares $\not\equiv 2\pmod5$

But by the Rational Root Test,  if $\,x^2\!-15\,$ has a rational root, it is an integer, contra above.

You can't directly use the linked proof because it uses $\,15\mid k^2\,\Rightarrow\, 15\mid k,\,$ which works because $\,15\,$ is squarefree. But $\,5n+2\,$ needn't be squarefree, e.g. it's $\,12\,$ for $\,n=2,\,$ and $\,12\mid 6^2\,$ but $\,12\nmid 6.\,$ But we can always reduce to squarefree radicands: write $\,5n+2 = kj^2.\,$ Then $\sqrt {5n+2} = j\sqrt k\,$ so $\,\sqrt{5n+1}\,$ is rational iff $\sqrt k\,$ is rational. By above $\,5n+2\,$ is not a square so it has squarefree part $\,k>1,\,$ so the linked classical method proves $\sqrt k$ irrational $\,\Rightarrow\sqrt{5n+1} = j\sqrt k\,$ irrational.

For the second, note that if $\ d\mid 5n\!+\!7,\, 3n\!+\!4\$ then

$${\rm mod}\ d\!:\,\ \dfrac{7}5 \equiv -n \equiv \dfrac{4}3 \,\Rightarrow\, 0\equiv 7\cdot 3 - 5\cdot 4\equiv 1\,\Rightarrow\, d\mid 1$$

Another way is to use Cramer's rule (or elimination) to solve for $\,n\,$ and $\,1\,$ below

$$\begin{eqnarray} \ 3\,n\, +\, 4\cdot 1 &=& i\\ \\ 5\ n\, +\, \ 7\cdot 1 &\ =\ & j\end{eqnarray} \quad\Rightarrow\quad \begin{array}\ n \ = \ \ \ \, 7 \,i\, -\, 4\ j \\\\ \color{#c00}{\bf 1}\ =\, {-5}\ \color{#0a0}i\, +\, 3 \ \color{#0a0}j \end{array}$$

Therefore, by the lower equation in the RHS system: $\ n\mid \color{#c00}{\bf 1}\,\$ if $\,\ n\mid \color{#0a0}{i,\,j}$

Remark $\$ In the same way we can prove more generally

Theorem $\$ If $\rm\,(x,y)\overset{A}\mapsto (X,Y)\,$ is linear then $\: \rm\gcd(x,y)\mid \gcd(X,Y)\mid \color{#c00}\Delta \gcd(x,y),\ \ \ \color{#c00}{\Delta := {\rm det}\, A}$

e.g.  in OP we have $\,\color{#c00}{\Delta =\bf 1}\,$ so the above yields $\ \gcd(3n+4,5n+7)\mid\color{#c00}{\bf 1}\cdot\gcd(n,1) = 1$

• Notice that $3^{-1}$ exists mod $d$ since $3$ is coprime to $d$, being coprime to its multiple $\,3n\!+\!4.\,$ Similarly we deduce that $\,5^{-1}$ exists, so both fractions are well defined. – Bill Dubuque Dec 5 '16 at 19:50

Since $$\{0,1,2,3,4\}\overset{x^2}{\longrightarrow}\{0,1,4\}\pmod{5}\tag{1}$$ and $$2\not\in\{0,1,4\}\pmod{5}\tag{2}$$ we know that $2$ is not a square mod $5$. That means that $$5n+2=x^2\tag{3}$$ has no solutions with $n,x\in\mathbb{Z}$.

If $\sqrt{5n+2}\in\mathbb{Q}$, then there is some $x\in\mathbb{Q}$ so that $$x^2-(5n+2)=0\tag{4}$$ which implies that $x$ is a rational algebraic integer, therefore, by this answer, it must be an integer. However, this contradicts that $(3)$ has no integer solutions.

Therefore, $$\sqrt{5n+2}\not\in\mathbb{Q}\tag{5}$$

Since $$3(5n+7)-5(3n+4)=1\tag{6}$$ we know that $5n+7$ and $3n+4$ have no common factors.