Prove that $\sqrt{5521}-1$ is an irrational number I'm stuck on a textbook problem where I'm asked to prove that $\sqrt{5521} − 1$ is an irrational number by contradiction and prime factorization. 
I tried following steps the teacher did in her class $10 + $$\sqrt{5}$ , and I made it to the end of the first steps but I couldn't get past that as I didn't know what to do after ending up with $5521b^2 = a^2$. For her example, she did four cases for when the remainder is $1$, $2$, $3$, and $4$ but I don't understand where she is getting these values from. My guess is she did every number up to but not including $5$, but the number I'm given is $5521$ and that seems like an excessive number of cases, which makes me believe that I'm not doing this right. I don't even know if the steps she did on her slides would also apply to any irrational number, specifically the one I'm asked to solve. 
Any help or push in the right direction would be appreciated.
 A: You're right on track! You have $5521b^2= a^2$. Now let's factor $5521$...and we're done! It's prime. Now you know $5521$ divides the left side (it's sitting right there), so $5521$ must divide the right side. But as I'm sure you know, a prime number divides a number squared if and only if it divides the number itself, so $5521$ divides $a$ (because it divides $a^2$). 
But then $5521$ times some number must give you $a$. Let's say $5521M=a$ for some integer $M$. Plug in this for $a$, do some cancelling, and see if you can get $5521$ to divide $b$, because then you have your contradiction. [After all, you assumed $a/b$ was in lowest form, so no number should divide them both, except maybe $\pm 1$.]
You're right on track! You have $5521b^2= a^2$. Now let's factor $5521$...and we're done! It's prime. Now you know $5521$ divides the left side (it's sitting right there), so $5521$ must divide the right side. But as I'm sure you know, a prime number divides a number squared if and only if it divides the number itself, so $5521$ divides $a$ (because it divides $a^2$). 
But then $5521$ times some number must give you $a$. Let's say $5521M=a$ for some integer $M$. Plug in this for $a$, do some cancelling, and see if you can get $5521$ to divide $b$, because then you have your contradiction. [After all, you assumed $a/b$ was in lowest form, so no number should divide them both, except maybe $\pm 1$.]
EDIT Thanks you the comment for pointing something out. Given the proof slides, they prove a prime divides a square if and only if it divides the number is given as a specific case example. But this is more general, and let's prove that fact now... [In fact, we will prove something better! If $p$ divides a product of two numbers $a,b$, i.e. if $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$.]
Lemma. If $p$ divides the product of two integers, $ab$, then $p$ divides $a$ or $p$ divides $b$.
Proof. Well, either $p$ divides $a$ or it doesn't. If $p$ divides $a$, we're done! Because then $p$ divides at least one of them. So we're left in the case that $p$ does not divide $a$. 
Well, then we can use the Euclidean Algorithm (which you probably talked about when learning about division), to find integers $x,y$ so that $xp+ya=1$. Now let's multiple by $b$. Then we have
$$
xbp + yab=b
$$
But $p$ divides $xbp$ (there's a $p$ right there!) and it divides $yab$ (because $p$ divides $ab$ by assumption). Then $p$ must divide the sum $xbp + yab$ because it divides each of the numbers $xbp,yab$ individually. But then $p$ divides the left side, so it must divide the right side, which is $b$. But then again, $p$ divides at least one of them! QED
But what does this have to do with your problem? Well, you have $p$ divides $a^2$. But $a^2$ is a product of two numbers, namely $a$ and itself. Because $p$ divides $a^2= a\cdot a$, it must divide one of them. But both of them are $a$! Therefore, $p$ divides $a$.
PS. This works the other way too. If $p$ divides $a$, then $a= Mp$ for some integer $M$. But then $a^2=(Mp)^2=M^2p^2=(M^2p)p$ so that $p$ divides $a^2$ also. Hence, $p$ divides $a^2$ if and only if $p$ divides $a$.
A: Well, it's relatively easy to prove that if $n\in\Bbb N$ isn't a perfect square (that is, the square of another natural number), then $\sqrt n$ is irrational. 
Now, if $\sqrt{5521}-1$ were rational,  so would be $\sqrt{5521}$.
A: Let, $\sqrt{5521}–1$= $\frac{p}{q}$, where $p$ and $q$ are co-prime integers.
 So,$\sqrt{5521}$=$\frac{p}{q}$+$1$
Squaring both side......
So, 5521=$\frac{p^2}{q^2}$+$1$+2$\frac{p}{q}$.
So, 5520=$\frac{p^2}{q^2}$+2$\frac{p}{q}$
Since, 5520 is integer...
So, $\frac{p^2}{q^2}$+2$\frac{p}{q}$must be an integer, which is not possible .
Hence, Contradiction!!
