Question 944729
There is more than one answer to that.
 
A very popular Pythagorean triple (and the only one I remember) is {{{"(3,4,5)"}}} .
A (right) triangle with those measurements can be scaled up to another triangle whose measurements are other Pythagorean triples.
with a scale factor of {{{5}}} , we get {{{highlight("(15,20,25)")}}} .
 
A Pythagorean triple like {{{"(15,20,25)"}}} is considered a "multiple", while {{{"(3,4,5)"}}} is considered a "primitive".
 
A formula to generate the primitive Pythagorean triples is 
{{{2mn}}}{{{","}}}{{{m^2-n^2}}}{{{","}}}{{{m^2+n^2}}} with positive integers {{{m}}} and {{{n}}} such that {{{m>n}}} .
 
We can use that formula to generate another primitive Pythagorean triple that includes {{{5}}} .
{{{m^2+n^2=5}}}<--->{{{system(m=2,n=1)}}} generates {{{system(2mn=2*2*1=4,m^2-n^2=2^2=1^2=4-1=3,m^2+n^2=2^2+1^2=4+1=5)}}} ,
which gives us the very popular {{{"(3,4,5)"}}} .
We cannot make {{{2mn=5}}} ,
but we can also make {{{m^2-n^2=5}}}<-->{{{(m+n)*(m-n)=5}}} with {{{system(m+n=5,m-n=1)}}} <---> {{{system(m=3,n=2)}}} .
{{{system(m=3,n=2)}}} gives us {{{system(2mn=2*3*2=12,m^2-n^2=3^2-2^2=9-4=5,m^2+n^2=3^2+2^2=9+4=13)}}} .
Then, primitive Pythagorean triple {{{"(5,12,13)"}}} can be scaled up by a factor of {{{5}}} to {{{highlight("(25,60,65)")}}} .
 
We can also use the formula to generate another primitive Pythagorean triple that includes {{{25}}} .
We cannot make {{{2mn=25}}} or {{{m^2-n=(m+n)(m-n)=25}}} .
We can only make {{{m^2+n^2=25}}} and that requires {{{system(m=4,n=3)}}} .
{{{m^2+n^2=25}}}<--->{{{system(m=4,n=3)}}} generates {{{system(2mn=2*4*3=24,m^2-n^2=4^2-3^2=16-9=7,m^2+n^2=4^2+3^2=16+9=25)}}} .
That gives us {{{highlight("(7,24,25)")}}} .