Question

find integer n if such that n-86 and both n+86 both are perfect squares​

Answer 1

Answer:

n is 1850

consider two numbers x and y such that

${x}^{2} = n + 86$

${y}^{2} = n - 86$

hence

$(x - y)(x + y) = (n + 86) - (n - 86)$

$(x - y)(x + y) = 172 = 2 \times 86$

Now we proceed via trial and error method

let's assume

$(x - y) = 2 \\ (x + y) = 86$

Solving we get

$x = 44 \\ y = 42$

Hence we have

${x}^{2} = 1936 = 1850 + 86 \\ {y}^{2} = 1764 = 1850 - 86$

Hence n is 1850

Answer 2

Answer:

n is 1850

consider two numbers x and y such that

{x}^{2} = n + 86x

2

=n+86

{y}^{2} = n - 86y

2

=n−86

hence

(x - y)(x + y) = (n + 86) - (n - 86)(x−y)(x+y)=(n+86)−(n−86)

(x - y)(x + y) = 172 = 2 \times 86(x−y)(x+y)=172=2×86

Now we proceed via trial and error method

let's assume

\begin{gathered}(x - y) = 2 \\ (x + y) = 86\end{gathered}

(x−y)=2

(x+y)=86

Solving we get

\begin{gathered}x = 44 \\ y = 42\end{gathered}

x=44

y=42

Hence we have

\begin{gathered}{x}^{2} = 1936 = 1850 + 86 \\ {y}^{2} = 1764 = 1850 - 86\end{gathered}

x

2

=1936=1850+86

y

2

=1764=1850−86

Step-by-step explanation:

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