If $(30X+7)(30Y+1) = 50437$, then what are the integer solutions? Any way to solve it without searching $50437$ factors?
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1Please note (again) that this kind of equations are not classified as functional equations but diophantine equations, hence my tag-editing. – Dmoreno Aug 15 '14 at 18:43
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1@Dmoreno: thank you for note – Kumar Aug 15 '14 at 18:46
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1I doubt you can find the integers solutions without finding the factors... But if you want them: they are $\lbrace \pm 1,\pm 31, \pm 1627, \pm 50437\rbrace$ – Darth Geek Aug 15 '14 at 18:46
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is there anyway without examine the factors of 50437? – Kumar Aug 15 '14 at 18:52
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1The search is short, since we are interested only in factors of the form $30X+7$ and $30Y+1$. – André Nicolas Aug 15 '14 at 18:54
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is there other way (no problem if it is long) – Kumar Aug 15 '14 at 18:58
3 Answers
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If you're looking for solutions where $X$ and $Y$ are integers, we see that $(30X + 7)$ and $(30Y + 1)$ must also be integers. Hence, you need only examine the factors of $50437$.
Ken
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If you are asking for a computationally easier way, then I doubt that's possible. This is the foundation of modern cryptography. Your question is equivalent to asking "Is there an easier way of finding integer factors". Only difference being the restriction on the form of factors - $30X+7$ and $30Y+1$ Please read:
http://en.wikipedia.org/wiki/Integer_factorization
However, if you just want to avoid the technicality of factorization, you can resolve the equation in terms of $Y$, plot the function and check for values at integer points and mark the integer solution.
$$Y = \frac{\frac{50437}{30X+7} - 1}{30} $$
Anything you do in the end is equivalent to factorization.
ste_kwr
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Your problem is to solve $xy=50437$, i.e. factor $50437$, into factors of special form. There are clever factorization methods, that are more efficient than just "searching for factors", but one way or another you are doing the factoring.
Conifold
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