4

find all positive integers $a,b,c$ such that

$abc=24$

$ab+bc+ca=38$

If particular values are given then we can easily find the solution but I am searching for some short general method. Is there any sufficient condition for the constant values to have integer solution?

Satvik Mashkaria
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  • For a single equation, the formula is simple. You can see it there. http://math.stackexchange.com/questions/419766/number-of-solution-for-xy-yz-zx-n/713998#713998 – individ Jan 06 '15 at 09:53

2 Answers2

1

Problems like this can be solved with a combinatorial approach. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. You have to put these in the 'boxes' a, b and c, as well as 'boxes' ab, ac and bc. Solutions can be made with 1,2 and 12. Hope this helps.

nilo de roock
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$ab+bc+ca=38 \Rightarrow a^2(b+c)+24=38a$. From the function $f(a)=a^2(b+c)-38a+24$ we get $38^2 \ge 4\cdot 24\cdot (b+c) \Rightarrow 15 \ge (b+c)$. You have now 13 values for b+c but not for every value will $38^2-4\cdot 24\cdot (b+c)$ be a perfect square and we also need $a_{1,2}=\frac{38 \pm \sqrt{38^2-4\cdot 24\cdot (b+c)}}{2(b+c)}$ to be an integer. We always get a and (b+c), but b and c can be calculated using $bc=\frac{24}a$. You will get (1,2,12) and his permutations as the only solution.

Note: we see from the solution that $a+b+c=15$ is always constant, so proving that would give us an even more simple solution bcs then we know that a,b,c are the solutions to the equasion $x^3-15x^2+38x-24=(x-1)(x-2)(x-12)=0$

This is as much as I know, if there is a faster solution that can covert them all please let me know.

CryoDrakon
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