Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference.
Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference.
Let the two natural numbers be $x$ and $y$.
According to the question:
$x^{2}+y^{2}=25(x+y) \ldots$ (i)
$x^{2}+y^{2}=50(x-y) \ldots$ (ii)
From (i) and (ii), we get:
$25(x+y)=50(x-y)$
$\Rightarrow x+y=2(x-y)$
$\Rightarrow x+y=2 x-2 y$
$\Rightarrow y+2 y=2 x-x$
$\Rightarrow 3 y=x \quad \ldots$ (iii)
From (ii) and (iii), we get:
$(3 y)^{2}+y^{2}=50(3 y-y)$
$\Rightarrow 9 y^{2}+y^{2}=100 y$
$\Rightarrow 10 y^{2}=100 y$
$\Rightarrow y=10$
From (iii), we have:
$3 \times 10=x$
$\Rightarrow 30=x$
Hence, the two natural numbers are 30 and 10 .
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