# elem.num.theo - diophantine equations

• Oct 9th 2008, 02:32 PM
cassiopeia1289
elem.num.theo - diophantine equations
define all solutions in integers to diophantine equations:

56x+72y=40

ok - I'll write my steps, my problem is that I end up with no solutions at all - maybe someone can find where I went wrong ... ?

SO:
72 = 56(1) + 16
56 = 16(3) + 8
16 = 8(2)
so gcd(56,72) = 8

next:
8 = 56 - (16)(3)
= 56 - 3(72-56)
= 56(4) + 72(-3)

40 = 56(20) + 72(-15)
so x0=20 & y0=-15

so:
x = 20 + 9t, for some int. t
y = -15 - 7t, for some int. t

and I get:
-(15/7) > t > -(20/9)
but there are no integers that work.
did I do something stupid in algebra or can there really be no solutions?
• Oct 9th 2008, 02:39 PM
icemanfan
Quote:

Originally Posted by cassiopeia1289
x = 20 + 9t, for some int. t
y = -15 - 7t, for some int. t

Stop right there. Those are your solutions, for all integers t.
• Oct 9th 2008, 02:44 PM
cassiopeia1289
but normally we find the integers of t
thats the what the question asks for: all solutions in the integers
and then use that to find the x&y's for each of the t's
• Oct 9th 2008, 02:47 PM
icemanfan
t can be any integer and

x = 20 + 9t
y = -15 - 7t

will be a solution.

So that is the solution set, and it is defined for all integers t. If you want to know what possible values of t are, the answer is all integers.
• Oct 9th 2008, 03:00 PM
cassiopeia1289
well - ok - what I would normally do (what we always do in class and on homework) is what I did above:

x = 20 + 9t, for some int. t
y = -15 - 7t, for some int. t

so you set 20 + 9t > 0
where t > -(20/9)
and then -15 - 7t > 0
where t < -(15/7)

and so you get
-(15/7) > t > -(20/9)

then (normally) from there we would take all the values of t in that restriction
(say for instance you had .5>t>2.5 - so t=1&2, and then you would plug those t values back into your x=blah and y=blah - and those are your solutions)

but with this problem - there are no integers in that restriction so that makes me wanna say no solutions ... ?
• Oct 9th 2008, 03:06 PM
Jhevon
Quote:

Originally Posted by cassiopeia1289
well - ok - what I would normally do (what we always do in class and on homework) is what I did above:

x = 20 + 9t, for some int. t
y = -15 - 7t, for some int. t

so you set 20 + 9t > 0
where t > -(20/9)
and then -15 - 7t > 0
where t < -(15/7)

and so you get
-(15/7) > t > -(20/9)

then (normally) from there we would take all the values of t in that restriction
(say for instance you had .5>t>2.5 - so t=1&2, and then you would plug those t values back into your x=blah and y=blah - and those are your solutions)

but with this problem - there are no integers in that restriction so that makes me wanna say no solutions ... ?

no, note that what you were solving for were the positive integer solutions. you wanted both x and y to be greater than zero. so you set the expressions for x and y to be greater than zero respectively, and then solve for the range that t has to be within. no such restriction was given here. we accept negative as well as positive solutions as far as this problem is concerned
• Oct 9th 2008, 03:23 PM
cassiopeia1289
ah - ok that makes sense - thank you for being so patient!