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I am not from pure math background. I am working on an algorithm which works good for all the practical reasons based on the following assumption.

that, if ab = cd and a+b = c+d then, either a = c & b = d or a = d & b = c. where, $a,b,c,d$ $\in$ $\mathbb{Z}$

I am not sure if this is true in the all conditions, but if it's can anyone provide proof, why ?

Thanks in advance.

Addition -1 : Is this also true for pair of N numbers ?

Addition -2 (Answer for the above) - I think I got it, @saulspatz's proof can be generalized (already sufficient), we can say,

After proof since, $a = c $ & $b = d$ then, by saying $b = b1 + b2$ and $c = c1 + c2$ proof can be further generalized for the addition and multiplication of pair of $N$ such numbers.

$a1, a2,...aN \in \mathbb{Z}$

$b1, b2,....bN \in \mathbb{Z} $

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  • $\begingroup$ What do you mean by "Is this also true for pair of N numbers ?"? What constraints do you have in mind? $\endgroup$
    – lightxbulb
    Commented Jan 14, 2019 at 19:14
  • $\begingroup$ in this case its just 2 numbers a + b = c+ d ... in case where its a1 + a2 + ... +an = b1 + b2 +b3 ...bn how can the proof be extended ? $\endgroup$
    – rahulb
    Commented Jan 15, 2019 at 1:15
  • $\begingroup$ I updated my answer to give an example with more values. Also can you explain what your algorithm aims to achieve? @rahulb $\endgroup$
    – lightxbulb
    Commented Jan 15, 2019 at 10:06

4 Answers 4

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Let $a+b=n=c+d$. Then $$ab=cd\implies a(n-a)=c(n-c)$$ so that $$an-a^2=cn-c^2,$$ or$$(a-c)n=a^2-c^2=(a-c)(a+c)$$

Either $a-c,$ and we are done, or $a-c\ne 0$ so we can cancel the $a-c$ to get $$n=a+c$$

Together with $n=aa+b$ this gives $b=c.$

In short, your assumption is correct.

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  • $\begingroup$ thanks @saulspatz .. in case where its a1 + a2 + ... +aN = b1 + b2 +b3+ ...+bN and a1* a2 * a3*... aN = b1 * b2 *b3 ...*bN how can the proof be extended ? $\endgroup$
    – rahulb
    Commented Jan 15, 2019 at 1:16
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Only an idea: if $$a+b=c+d$$ we get by squaring $$a^2+b^2=c^2+d^2$$ since $$ab=cd$$ so we get $$a^2-c^2=d^2-b^2$$ or $$a^2-d^2=b^2-c^2$$

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This is true.

In general, $r$ and $s$ are the two roots of the quadratic equation $$ x^2 -(r+s)x + rs = (x-r)(x-s). $$ so knowing the sum and the product determines the values (up to swapping them).

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Let $\exists \delta_1, \delta_2 \in \mathbb{Z}$ such that $c = a + \delta_1, d = b + \delta_2$ and $(\delta_1,\delta_2) \neq (0,0)$. However from $a+b=c+d$ it follows that $\delta_1 = -\delta_2$. Now $ab = cd = (a+\delta_1)(b-\delta_1)$, then $\delta_1^2 + (a-b)\delta_1 = \delta_1(\delta_1+a-b) = 0$. Since we assumed $\delta_1 \neq 0$ it follows that $\delta_1 = b-a$, however then we get $c = b, d=a$. Thus it necessarily holds that for the given constraints either $(a,b) = (c,d)$ which corresponds to $\delta_1 = 0$, or $(a,b)=(d,c)$ which corresponds to $\delta_1 = b-a$. Note that this is true for $a,b,c,d \in \mathbb{R}$.

Edit: Since the original question was extended through an edit, I'll address that also. If you have more variables you need more equations, for example: $$a,b,c,d,e,f \in \mathbb{R}$$ $$a+b+c = d+e+f$$ $$ab + bc + ac = de + ef + df$$ $$abc=def$$

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