Number of Integral values satisfying $17^2+n^4=m^2$. I got this question in my exam and I couldn't solve this.
The only thing I feel I could infer from this was that it has something to do with the sides of a right angled triangle. Could anyone deconstruct this for me?
 A: Hint: $17^2 = m^2 - n^4$ can be another way of looking at it which can be factorized using $a^2 - b^2 = (a+b)(a-b)$ 
And since m and n are integers, $(m+n^2)$ and $(m-n^2)$ are both integers as well as factors of $17^2$.
Can you carry on from here?
A: The tree of primitive Pythagorean triples gives us $17^2+144^2=145^2$, where $144^2=12^4$. Of course there is the trivial solution with $n=0$. On the other hand $17^2=(m-n^2)(m+n^2)$ shows that these are all solutions, up to sign.
A: We have:
$$17^2=m^2-n^4=m^2-(n^2)^2=(m-n^2)(m+n^2)$$
We can factorize $17^2$ only as the product of $1$ and $289$, or $17$ with itself. In the first case:
$$m-n^2=1, m+n^2=289 \implies m=145,n^2=144 \implies (m,n)=(145,12)$$
In the other case, we obviously have $n^2=0$ which gives $m=17$. Since $m$ and $n$ can also be negative, the solutions are:
$$(m,n)=(\pm 145, \pm 12), (\pm 17, 0)$$
A: If we solve side-A of Euclid's formula for $n$, we get a a function that will let us find all Pythagorean triples, if they exist, for that value of $A$ with a small finite search of values of $m$. Any $f(A,m)$ that generates an integer $n$ yields the $m,n$ pair for a corresponding triple.
$$\mathbf{A=m^2-n^2\Rightarrow n=\sqrt{m^2-A}\qquad\qquad  \lceil\sqrt{A+1}\rceil \le m \le \biggl\lceil\frac{A}{2}\biggr\rceil}$$
The lower limit ensures $m^2>A$ and the upper limit ensures $m-n\ge 1$.
$$A=17\Rightarrow \lceil\sqrt{17+1}\space\rceil=5 \le m \le \biggl\lceil\frac{17}{2}\biggr\rceil =9\quad \quad f(17,9)\Rightarrow n=8\quad F(9,8)=(17,144,145)$$
As indicated in other posts, $144=12^2$ so (using $A^2+B^4=C^2$) we  have $17^2+12^4=145^2$.  Also, since $A$ is prime, meaning the triple is primitive, we have no other solutions.
