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I have seen few math problems online, about solving one equation with two unknowns, (which is not possible as the number of equations should match the number of unknowns), but I thought is there any way to prove that LHS = RHS for some integer (...,-1,0,1,2...) numbers without substituting values like in trail and error method.

Suppose, I have an equation say 3x + 5y = 22, by looking at it for few seconds, I can substitute the values x = 4 and y = 2, but I want to know if there is any way to know that there exist a solution for the equation such that x and y are integers.

P.S. I don't know what to tag please edit the tags if found any better ones.

YouKnowWhoIAm
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  • By "real and non-fractional", I assume you mean integers, i.e., the set $\mathbb{Z}$, or perhaps the natural numbers, i.e., the set $\mathbb{N}$, possibly with or without including $0$, depending on the definition you may be using. Please explicitly & unambiguosly state what you mean in the question text. Thanks. – John Omielan Sep 15 '19 at 05:36
  • Yes, I forgot the word, it is integers, and it includes 0, thanks – YouKnowWhoIAm Sep 15 '19 at 05:40
  • Integers always includes $0$. Do you mean natural numbers, including $0$ instead, i.e., integers which are non-negative? – John Omielan Sep 15 '19 at 05:41
  • Pure integers, from -Inf to +Inf, or the integer limit – YouKnowWhoIAm Sep 15 '19 at 05:42
  • I don't understand your comment above. What "integer limit" are you referring to? From your updated text saying "... for some integer (0,1,2...) ...", it seems you mean "non-negative integers". You may wish to use this phrase as it's clear & unambiguous. – John Omielan Sep 15 '19 at 05:44
  • Now with your latest update, it seems you mean all integers instead. Although it doesn't hurt, note you don't to state "(...,-1,0,1,2...)" as you can reasonably assume anybody reading it already knows what integers are. – John Omielan Sep 15 '19 at 05:46
  • No, I mean any number that is not of the p/q form, can be negative, positive or zero.(...,-2,-1,0,1,...). I am not an English speaker, which might have caused the ambiguity, because I have every number in the mind but don't know the name. – YouKnowWhoIAm Sep 15 '19 at 05:46
  • Let us continue this discussion in chat. – YouKnowWhoIAm Sep 15 '19 at 05:49
  • No worries about not knowing the name. It's just integers, as I mention above. As for continuing the discussion in chat, I don't think there's anything more to discuss. Also, it is just before $11$ p.m. my time, and I am already late going to bed as I need to get up early tomorrow, so I don't have time to participate in chat right now. However, if you're still not clear on something related to this, perhaps we can chat sometime tomorrow instead? – John Omielan Sep 15 '19 at 05:51
  • Now that the ambiguity is cleared, can you provide me a solution. – YouKnowWhoIAm Sep 15 '19 at 05:52
  • Not right now since, as I wrote, I'm going to bed. However, if nobody else has satisfactorily answered your specific question when I check tomorrow & have some time, I will then see about answering it myself. – John Omielan Sep 15 '19 at 05:56

2 Answers2

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The slope of your line is $-\frac{3}{5}$. This is to say that for every $5$ units you travel in the $x$-direction, you move $-3$ units in the y-direction.

Since you know that $x=4, y=2$ is a solution, then you can move right $5$ units and down $3$ units to get another solution. You can do this infinitely many times. In addition, you can go in the opposite direction (left and up) and get infinitely many integer points as well.

Then the general solution where both $x$ and $y$ are integers is:

$$(4+5t, 2-3t)$$

where $t$ can be any integer value.

Toby Mak
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  • Since here I know there is a solution, you can give me (4+5t, 2-3t) as a general solution, but what if for some other arbitrary equation, 8x+3y=12, here with the same trial and error approach we know that there is no solution possible with integer values. I want to know whether there is any method to say that an equation has integer solution – YouKnowWhoIAm Sep 16 '19 at 07:32
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    Obviously you have $(4,0)$ is a solution. For the general case, you can use the Euclidean algorithm. Once you use it to find the greatest common factor, you can work backwards through the steps to express say, $1$ in terms of multiples of $8$ and $3$. Try using this page to find one solution to $8x-3y=1$ and then apply the method in my answer. – Toby Mak Sep 16 '19 at 09:59
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    And also, unless you have something impossible (such as $2x+2y=3$), most equations will yield an infinite number of integer solutions. – Toby Mak Sep 16 '19 at 09:59
  • Correction: $(0,4)$ is a solution. – Toby Mak Sep 21 '19 at 09:09
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You can write $$y=-\frac{3}{5}x+\frac{22}{5}$$ and you will get the solutions $$(x,y)=\left(x,\frac{22}{5}-\frac{3}{5}x\right)$$

Toby Mak
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Dr. Sonnhard Graubner
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  • I added few points, I didn't write them properly, sorry for that, I want the numbers to be whole and non-fractional ones. – YouKnowWhoIAm Sep 15 '19 at 05:28