Solving a Diophantine equation?

63* #63*x+70*y+75*z=91#

2 Answers
Nov 28, 2017

#{ (x = 5k+2), (y = -12k-15h-8), (z = 7k+14h+7) :}#

for any integers #k# and #h#

Explanation:

Given:

#63x+70y+75z=91#

First note that all terms are divisible by #7# except #75z#.

So we require #z=7w# for some #w#.

Then:

#9x+10y+75w=13#

Note that #10y+75w# is divisible by #5#.

Hence we require #9x-13# to be divisible by #5#.

Hence #x=5k+2# for some #k#

Then:

#45k+18+10y+75w=13#

So:

#9k+2y+15w+1=0#

Since #2y+1# is odd, we require exactly one of #k# and #w# to be odd and the other even. We can express this by putting #w=k+2h+1#

Then:

#0 = 9k+2y+15(k+2h+1)+1 = 24k+2y+30h+16#

So:

#12k+y+15h+8 = 0#

Then note that #12k# and #15h# are both divisible by #3#, so we require #y = 3f-8# for some integer #f#.

Then:

#4k+f+5h = 0#

So for any integers #k# and #h#, we can put #f = -4k-5h#.

Then:

#y = 3f-8 = 3(-4k-5h-8) = -12k-15h-8#

and we have a solution of the original equation, with:

#{ (x = 5k+2), (y = -12k-15h-8), (z = 7k+14h+7) :}#

Nov 30, 2017

See below.

Explanation:

#63*x+70*y+75*z=91#

This equation can be solved using a technique inspired on the sieve of Eratosthenes.

1) #x + y + 7/63 y + z + 12/63 z = 91/63#

now calling #a = 7/63 y + 12/63 z - 91/63# we have

#color(red)(x+y+z + a=0)# and

#63 a= 7 y + 12 z- 91# and now

2) #9a = y + z + 5/7z - 91/7#

now calling #b = 5/7z - 91/7# we have

#color(red)(9a=y+z+b)# and now

3) # 7b=5z-91 rArr b+2/5b=z-91/5#

and now calling #c = 2/5b+91/5# we have

#color(red)(z=b+c)# and now

4) #5c=2b+91 rArr 2c+1/2 c=b +91/2# and calling

#d = 1/2c-91/2# we have #2c-b+d=0# and

#color(red)(c = 2d+91)#

but #z = b+c rArr z = 2c+d+2d+91 = 7d+3 xx 91#

now #y = 9a-(z+b) = 9a-12d-5 xx 91# and finally

#x = -(y+z+a) = 5d-10a+2 xx 91#

Resuming

#color(red)({(x =5d-10a+2 xx 91 ),(y = 9a-12d-5 xx 91 ),(z =7d+3 xx 91 ):}#