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204 SPECIAL TOPICS IN THEORETICAL ARITHMETIC

Here we have # + 1 equations with the # 4+ 2 unknowns
X, V1, ¥2, V3, * * *» Yn, 2. We may expect therfor an infinit num-
ber of solutions.

489. For any particular value of # this problem may be
solved by the method of § 417.

For example, if n = 5, setting s, ¢, %, v, w for y1, V2, V3, V4, Vs,
we have the following 6 equations with 7 unknowns:

x=15s 41
4s =5t + 1
4 = 5u + 1
4y = 50 + 1
49 = 5w+ 1
4w = 53
From the last equation 5z > 4
But 5114
Hence 234
Therfor z=4m
and hence w = Sm

Substituting this value for w in the preceding equation,
using the method of § 417 and going backward in the series
of equations, we find after considerable labor that the com-
plete solution of the equations is

x = 15,625¢ + 3121
= 3,125¢ + 624
2,500¢ + 499
2,000g + 399
1,600 + 319
1,280 + 255
= 1,024 + 204

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The solution of the problem is then easily completed.

490. We will now prepare to solve the problem for any
value of n.