Question
Find two consecutive positive integers, sum of whose squares is 365.
✓
Answer
Let the two consecutive positive integers be $x$ and $x+1$ According to the question
$x^2+(x+1)^2=365$
$\Longrightarrow x^2+x^2+2 x+1=365$
$\Longrightarrow 2 x^2+2 x-364=0$
$\Longrightarrow 2\left(x^2+x-182\right)=0 \text { or } x^2+x-182=0$
$\Longrightarrow x^2+14 x-13 x-182=0$
$\Longrightarrow x(x+14)-13(x+14)=0$
$\Longrightarrow(x-13)(x+14)=0$
Either $x-13=0$ or $x+14=0$
$\Longrightarrow x=13,-14$
Since the numbers are positive. so $x=-14$ is rejected.
Hence the required consecutive positive integers are $13,13+1=14$.
$x^2+(x+1)^2=365$
$\Longrightarrow x^2+x^2+2 x+1=365$
$\Longrightarrow 2 x^2+2 x-364=0$
$\Longrightarrow 2\left(x^2+x-182\right)=0 \text { or } x^2+x-182=0$
$\Longrightarrow x^2+14 x-13 x-182=0$
$\Longrightarrow x(x+14)-13(x+14)=0$
$\Longrightarrow(x-13)(x+14)=0$
Either $x-13=0$ or $x+14=0$
$\Longrightarrow x=13,-14$
Since the numbers are positive. so $x=-14$ is rejected.
Hence the required consecutive positive integers are $13,13+1=14$.
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