$\begingroup$I need to find three numbers x, y, z where: 1) the multiplication of all these numbers equals 36 2) the sum of these three equals to the square of the sum of the two. The question goes if there enough facts to find these three numbers and if so what are they?
Ok, the first one is clearly translated as x * y * z = 36. I am having some trouble translating the second one, the first part is x + y + z but should the second be = (x + y)^2, = (x + z)^2 or = (y + z)^2? And if so, how do I continue from there...
2 Answers
$\begingroup$$$\large 36=1.2^2.3^2$$ Now wlog: $$\large xyz=36\text{ and }x+y+z=(x+y)^2$$ Or: $$\large z=(x+y)^2-(x+y)=(x+y)(x+y-1)$$ So: $$\large36=xyz=xy(x+y)(x+y-1)$$ Now, since $x,y$ are natural numbers, you need to find two consecutive factors of $36$ that fit the data, which are: $(1,2),(2,3),(3,4)$. Out of which $(3,4)$ fits the data with $x=1,y=3$(wlog) and $x+y-1=3,x+y=4$ So, the numbers are: $$\large 1,3,12\\\large 1.3.12=36\quad\mid\quad 1+3+12=16=4^2=(1+3)^2$$
$\endgroup$$\begingroup$The second equation should be $$x+y+z=(x+y)^2$$ You just pick two of them to square. I can't think of a good way to solve this, but, assuming they are all integers, you can quickly go through the factorizations of $36$
$(1,1,36),(1,2,18)$,etc
and look for a set that sum up to a perfect square. It doesn't take too long to get to the three factors that will work.
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