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Let's assume we have:

a = 15
b = 10
number_of_squares = 12

Meaning:

Overall = 150
one_unit_overall = 150/12 = 12.5
one_unit_a = square root of 12.5 = 3.53

This means if I cut rectangle on $12$ equal pieces, $4 \cdot 3$ for example, the width and height of square will not fit in $a \cdot b$ rectangle. What am I mising?

I would appreciate it.

I get:

a_square = 14.12 != a_rectangle (15)
b_square = 10.59 != b_rectangle (10)
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  • 1
    $\begingroup$ You have solved $$12x^2=150\to x=\sqrt{12.5}$$ but another condition is $$nx=10;\;mx=15;\;mn=12$$ $x$ must be integer $\endgroup$ – Raffaele Nov 17 '17 at 10:51
  • $\begingroup$ @Raffaele This can not be solved if x is not integer? $\endgroup$ – Testing man Nov 17 '17 at 10:53
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    $\begingroup$ To rephrase what Raffaele said in more geometrical terms, if a "square tiling" exists, you solved for the side length of the tiling square (what Raffaele called $x$, what you denoted one_unit_a). The other condition given by Raffaele is about how many squares you have to line up horizontally/vertically to actually tile the rectangle: $n$ would be the number of squares in a column (if $b=10$ is the height) and $m$ the number per line (if $a=15$ is the width). To achieve what you want, you need $n,m$ to exist and be integers, or there is no solution. $\endgroup$ – N.Bach Nov 17 '17 at 12:13
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    $\begingroup$ The constraint $n,m$ integers, and the equalities $$nx=10;\ mx=15$$ implies that $x$ must at least be rational. Problem here is that $$x=\sqrt{\frac{25}2}=\frac 5{\sqrt 2}$$ is not rational (because $\sqrt 2$ is not). $\endgroup$ – N.Bach Nov 17 '17 at 12:17

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