Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Its is my assumption that they are not unique although am yet to find an arguement to prove this

share|cite|improve this question
Your first impulse, when thinking that something might be so, should be to try out a simple example. Like $a=2$ and $b=5$. Then see whether the single equation $2x+5y=1$ has a unique solution in integers. Examples should come before everything else in gaining understanding. – Lubin Feb 19 '13 at 21:46

Let $d=\gcd(a,b)$. Let $a=da'$ and $b=db'$.

Suppose that $ax_0+by_0=d$. Then the solutions of $ax+by=d$ are given by $$x=x_0-tb',\qquad y=y_0+ta',$$ where $t$ ranges over the integers.

Remark: It is easy to verify that these are solutions, since $(a)(-tb')+(b)(ta')=0$.

To prove that there are no others, suppose that $ax+by=d$. Then by subtraction we find that $a(x-x_0)+b(y-y_0)=0$, and therefore $a'(x-x_0)=-b'(y-y_0)$. Since $a'$ and $b'$ are relatively prime, we conclude that $a'$ divides $y-y_0$. Let $y-y_0=ta'$. Then $y=y_0+ta'$. It follows that $x=x_0-tb'$.

share|cite|improve this answer

Hint $\ $ Like any linear equation, if the associated homogeneous equation has nonzero solutions then the solutions to the non-homogeneous equation are not unique. This applies here since its homogenous form $\rm\ A\, X = (a,b)\cdot (x,y) = ax + by = 0\ $ has obvious solution $\rm\,X = (-b,a).$

Generally, for $\rm A\:$ linear, $\rm\ A\,X_1\! = B = A\,X_2 \ \iff\ 0 \:=\: A\,X_1\! - A\,X_2 = A\,(X_1\!-X_2)$

This implies that the general solution of $\rm\,\ A\,X = B\,\ $ is the sum of any fixed particular solution plus any solution of the associated homogeneous equation $\rm\ A\,X = 0.\:$ This property holds true for every linear operator, e.g. for matrices, linear differential equations, linear recurrences, etc, a fact which will come to the fore if you study linear algebra and vector/affine spaces, modules, etc.

share|cite|improve this answer

Uniqueness is dismissed by simply observing that $$(a+y)x+(b-x)y=ax+by.$$

share|cite|improve this answer
I like this answer best because it is nice and short and answers exactly the question posed. – Tara B Feb 19 '13 at 22:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.