Find two triangles of longest side length 25? I'm using the quadratic Diophantine equations to solve for two integer triangles of longest side $25$. It's been shown that for $a^2+b^2=c^2$, which goes to $x^2+y^2=1$ where $x=\frac ac, y=\frac bc, a=t^2−1, b=2t, c=t^2+1$. 
If I want to solve for $c=25$, how will I go about doing this? 
Just letting $c=25$ gives me an irrational answer for $b$ after having solved for $t$. I'm a little confused.
Thanks for your help.

All Pythagorean triples, i.e. triplets of positive integers such that $a^2+b^2=c^2$, can be expressed as $a=k(m^2-n^2)$, $b=2kmn$, $c=k(m^2+n^2)$; where $m$ and $n$ are coprime, $m>n$ and $m$ and $n$ have opposite parity. (Up to interchanging of $a$ and $b$.)
So in your case you want $k(m^2+n^2)=25$. This gives you the possibilities:
$k=1$ and $m^2+n^2=25$
$k=5$ and $m^2+n^2=5$
$k=25$ and $m^2+n^2=1$
For such small numbers, it is easy to find all expressions as a sum of two squares by hand, you will get $25=0^2+5^2=3^2+4^2$, $5=1^2+2^2$ and $1=1^2+0^2$.
Since you are interested only in non-zero values, you get only two possibilities for $(k,m,n)$ namely $(1,4,3)$ and $(5,2,1)$. From the first one you get $a=7$, $b=24$, $c=25$. The second one gives you $a=15$, $b=20$, $c=25$.
