I never heard of Euler Algebra but I know what you are talking about.

One way is countinued fractions, but I will use the backwards way of Euclidean Algorithm.

We calculate,

gcd(21,31)

31=1*21+10

21= 2*10+1

10= 9*1+0

Thus,

gcd(21,31)=1

Working backwards,

21-2*10=1

21-2(31-1*21)=1

Thus,

21-2*31+2*21=1

Thus,

21(3)+31(-2)=1

Thus,

21(5310)+31(-3540)=1770

So,

x=5310 and y=-3540

Is one solution of,

21x+31y=1770

Thus, all solutions are:

x=5310+31t

y=-3540-21t

We require that,

x,y>0

Thus,

5310+31t>0

-3540-21t>0

Solving both of these inequalities we get,

t>-171.2

t<-168.5

Thus,

-171.2<t<-168.6

Since t is in integer we have,

t=-171,-170,-169

Corresponding to 3 solution of x and y which are:

(x,y)=(9,51)

(x,y)=(40,30)

(x,y)=(71,9)