Find all solutions of the Diophantine equation 1245x-365y = 4567
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please use the example 1 as your reference
Transcribed Image Text:Find all solutions of the Diophantine equation 1245x-365y=4567
![Example 1. Find all possible integral solutions of 18x + 7y=302.
Solution: Since (18, 7) = 1 and 1 | 302, the given equation has solutions
Now, we solve for xo, Yo such that 18 xo +7 yo = 1
18=7 (2) + 4
7=4 (1) +3
4= 3(1) + 1
3=3(1)
Thus we have,
1=4-3
1-18-7(2)-[7-(18-7(2)}]
1=18 (2)-7 (5)
This means,
Xo = 2 and
Yo=5
. We substitute these values in the formula. Next. We find all values of t satisfying x>0 and y> 0.
604 +7t> 0
and
-1510-18t> 0
t<-83.9
t>-86.3
thus, t= -86, -85, -84
The table below presents the values of x and y for these values of t using
and
x = 604 + 7t
y=-1510-18t
t
y
-86
2
-85
9
-84
16
38
20
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0ae6d8fa-deba-4fae-adeb-e21f8c2c4edd%2Fd1c27d89-1a1f-43dd-95a5-379a1bb54671%2Fyahmjjl_processed.png&w=3840&q=75)
Transcribed Image Text:Example 1. Find all possible integral solutions of 18x + 7y=302.
Solution: Since (18, 7) = 1 and 1 | 302, the given equation has solutions
Now, we solve for xo, Yo such that 18 xo +7 yo = 1
18=7 (2) + 4
7=4 (1) +3
4= 3(1) + 1
3=3(1)
Thus we have,
1=4-3
1-18-7(2)-[7-(18-7(2)}]
1=18 (2)-7 (5)
This means,
Xo = 2 and
Yo=5
. We substitute these values in the formula. Next. We find all values of t satisfying x>0 and y> 0.
604 +7t> 0
and
-1510-18t> 0
t<-83.9
t>-86.3
thus, t= -86, -85, -84
The table below presents the values of x and y for these values of t using
and
x = 604 + 7t
y=-1510-18t
t
y
-86
2
-85
9
-84
16
38
20
2
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