INDETERMINATE EQUATIONS OF FIRST DEGREE 167

Of the two coefficients here, 7 and 15, we choose 7 as a
modulus.

15m + 4> 7
15m = — 4 (mod 7)
15 =1 (mod 7)
— 4 = 3 (mod 7)
m = 3 (mod 7)
m="1Tn + 3

Since the coefficient of m is 1, we can get the values of y
and x by substitution.

7y =15n4+4 =15 X Tn 4 45 + 4
y=1n+4+7

152 = 22y + 41 = 22 X 15# + 195
x = 22n + 13

Our complete solution is therfor (13 + 22%, 7 + 15n)

For x and y to be positiv we find that » must be greater
than — 1.

We have the following small table of values:

 

n x y
3 79 52

2 57 37

1 D 22

0 13 7
-1 — 9 - 8
—2 =31 — 23
-3 —'53 — 38

In this problem we have dealt successivly with the equations

15x — 22y = 41
15m — 7y = — 4
m— Tn= 33

It is interesting to note how the coefficients and unknowns
change or do not change here from equation to equation.