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Find all ordered pairs (x,y) of positive integers such that x^4 = y^2 + 71.

 May 5, 2020
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Find all ordered pairs (x,y) of positive integers such that \(x^4 = y^2 + 71\).

 

\(\begin{array}{|rcll|} \hline \mathbf{x^4} &=& \mathbf{y^2 + 71} \\ x^4-y^2 &=& 71 \\ (x^2-y)(x^2+y) &=& 71 \quad | \quad 71~ \text{is a prime number} \\ \hline \end{array} \)

 

\(\begin{array}{|rccc|} \hline 71= &(x^2-y)&\times&(x^2+y)\\ \hline 71= & 1 & \times & 71 \\ 71= & 71 & \times & 1 \\ \hline \end{array} \)

 

1.

\(\begin{array}{|lrclcrcl|} \hline (1)&\mathbf{x^2-y} &=& \mathbf{1} &\text{or}& x^2 &=& 1+y \\ (2)&\mathbf{x^2+y} &=& \mathbf{71} \\ &(1+y)+y &=& 71 \\ &1+2y &=& 71 \\ &2y &=& 70 \\ &\mathbf{y} &=& \mathbf{35} \\\\ &x^2 &=& 1+y \\ &x^2 &=& 1+35 \\ &x^2 &=& 36 \\ &\mathbf{x} &=& \mathbf{\pm 6}\\ \hline \end{array} \)

 

2.

\(\begin{array}{|lrclcrcl|} \hline (1)&\mathbf{x^2-y} &=& \mathbf{71} &\text{or}& x^2 &=& 71+y \\ (2)&\mathbf{x^2+y} &=& \mathbf{1} \\ &(71+y)+y &=& 1 \\ &71+2y &=& 1 \\ &2y &=& -70 \\ &\mathbf{y} &=& \mathbf{-35} \\\\ &x^2 &=& 71+y \\ &x^2 &=& 71-35 \\ &x^2 &=& 36 \\ &\mathbf{x} &=& \mathbf{\pm 6}\\ \hline \end{array}\)

 

\((x,y)= (-6,35)~(6,35)~(-6,-35)~(6,-35)\)

 

laugh

 May 5, 2020

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