Find-all-the-triples-of-positive-integers-x-y-z-so-that-x-y-2020-y-z-2020-is-a-rational-number-and-x-2-y-2-z-2-be-a-prime-number-

Question Number 97554 by 1549442205 last updated on 08/Jun/20

Find all the triples of positive integers (x; y; z)

so that \frac{x - y \sqrt{2020}}{y - z \sqrt{2020}} is a rational number

and x^{2} + y^{2} + z^{2} be a prime number .

Commented by Rasheed.Sindhi last updated on 09/Jun/20

\frac{x - y \sqrt{2020}}{y - z \sqrt{2020}} = \frac{x - 2 y \sqrt{505}}{y - 2 z \sqrt{505}}

= \frac{x - 2 y \sqrt{505}}{y - 2 z \sqrt{505}} \times \frac{y + 2 z \sqrt{505}}{y + 2 z \sqrt{505}}

= \frac{xy - 2020 yz + 2 xz \sqrt{505} - {2 y}^{2} \sqrt{505}}{y^{2} - {2020 z}^{2}}

= \frac{xy - 2020 yz + 2 \sqrt{505} (xz - y^{2})}{y^{2} - {2020 z}^{2}}

\Rightarrow xz - y^{2} = \sqrt{505} (I m p o s s i b l e ∵ x, y, z \in Z^{+})

Commented by prakash jain last updated on 08/Jun/20

or x z - y^{2} = 0

Commented by Rasheed.Sindhi last updated on 08/Jun/20

T h a n k y o u s i r!

Commented by Rasheed.Sindhi last updated on 09/Jun/20

I f x z - y^{2} = 0

y^{2} = x z

x^{2} + y^{2} + z^{2} = x^{2} + x z + z^{2}

= {(x + z)}^{2} - x z = {(x + z)}^{2} - y^{2}

(x + y + z) (x - y + z) \in P

\Rightarrow

x + y + z = 1 \land x - y + z \in P

(B u t x + y + z \neq 1 f o r x, y, z \in Z^{+})

o r

x - y + z = 1 \land x + y + z \in P

x - y + z = 1 \Rightarrow

x = y \land z = 1 ∣ y = z \land x = 1

L e t x = y = k \Rightarrow x + y + z = 2 k + 1

(S i m i l a r l y y = z = k \Rightarrow x + y + z = 2 k + 1)

\dots

x = z = 1

\dots .

Commented by prakash jain last updated on 09/Jun/20

IF

x - y + z = 1

y = x + z - 1

y^{2} = x z

{(x + z - 1)}^{2} = x z

x^{2} + 2 x (z - 1) + 1 = x z

x^{2} + x (z - 2) + {(z - 1)}^{2} = 0

{(z - 2)}^{2} - 4 {(z - 1)}^{2} is whole square

((z - 2) + 2 (z - 1)) (z - 2 - 2 z + 2)

= - z (3 z - 4)

z = 1 is only option

x^{2} - x = 0

\Rightarrow x = 1, y = 1

so 1, 1, 1 is the only solution .

Answered by 1549442205 last updated on 09/Jun/20

Let \frac{x - y \sqrt{2020}}{y - z \sqrt{2020}} = \frac{m}{n}, where m, n \in Z, \gcd (m; n) = 1

\Leftrightarrow nx - 2 ny \sqrt{505} = my - 2 mz \sqrt{505}

\Leftrightarrow (2 mz - 2 ny) \sqrt{505} = my - nx

if 2 mz - 2 ny \neq 0 then \sqrt{505} = \frac{my - nx}{2 mz - 2 ny} \in Q (set of rational numbers)

This is a contradiction because \sqrt{505} be irrational number

Hence, we need must have {\begin{cases} 2 mz - 2 ny = 0 \\ my - nx = 0 \end{cases}

it follows that \frac{m}{n} = \frac{x}{y} = \frac{y}{z} \Rightarrow y^{2} = xz . Now we

denote \gcd (x, z) = d \Rightarrow {\begin{cases} x = ad \\ z = bd \end{cases} (1) with a, b \in N and \gcd (a, b) = 1

\Rightarrow y^{2} = {abd}^{2} \Rightarrow a = u^{2}, b = v^{2} (u, v \in N^{*}) \Rightarrow y = uvd . Replace into

(1) we get x = u^{2} d, z = v^{2} d .

From this we get x^{2} + y^{2} + z^{2} = d^{2} (u^{4} + u^{2} v^{2} + v^{4})

= d^{2} [{(u^{2} + v^{2})}^{2} - u^{2} v^{2}] = d^{2} (u^{2} - uv + v^{2}) (u^{2} + uv + v^{2})

From the condition x^{2} + y^{2} + z^{2} is a prime we

infer that {\begin{cases} d^{2} = 1 \\ u^{2} - uv + v^{2} = 1 \end{cases} . Therefore,

{(u - v)}^{2} = 1 - uv ⩾ 0 \overset{u, v \in N^{*}}{\Leftrightarrow} u = v = 1 \Leftrightarrow x = y = z = 1

Thus, (x; y; z) = (1; 1; 1) is unique triple satisfying

requirement of the problem