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$x^2 = yz,$ $x + y + z = 61$

positive integral solutions.

On wolframalpha, i could find the solution, but not how to solve. I tried by taking a single equation $xz + x^2 +z^2 - 61z = 0$ Tried solving for both x & z separately but did not find a way.

Please help.

EDIT: This is a high school problem in numerical analysis. Diophantine Equations is not yet in course.

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  • $\begingroup$ Thanks, Took a little time but this is much better.... $\endgroup$ – Shh Apr 6 '18 at 7:03
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Suppose that $y$ and $z$ are not co-primes. In that case there must be a prime number p such that $p|y$ and $p|z$. From the first equation we get that $p|x$ and therefore $p|(x+y+z)$ or $p|61$ which is impossible (61 is a prime).

In other words $y$ and $z$ are co-primes. But their product is a perfect square and that is possible only if both numbers are squares. Which means that possible values for $y,z$ are:

$$y,z \in \{1, 4, 9, 16, 25, 36, 49\}$$

...which means that you have just a few pairs to check manualy and most of them can be easily discarded. Also note that at least one of them must be odd (otherwise $x,y,z$ would be all even which is impossible).

For example, 49 can be paired only with 9, 4, 1 (no solution there). 36 can be paired only with 1 and 9 (no solution). 25 can be paired only with 25, 16, 9, 4, 1. We have one solution there ($y=25, z=16, x=20$). Values 16,9,4,1 are just too small (less than 61/3) and should not be checked at all.

So we have exactly two solutions: $y=25, z=16, x=20$ and $y=16, z=25, x=20$

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  • $\begingroup$ Did not even think about the concept of co primes. Perfect square I thought of but could not eliminate other combinations with co primes. Thank You $\endgroup$ – Shh Apr 6 '18 at 8:20
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I wanted to see that a more algebraic approach looked like $$\begin{align} &\left[x^2=yz \quad | \quad z=61-x-y\right]\implies x^2=61y-xy-y^2 \\ &\qquad x^2+xy+\frac{y^2}{4}-\frac{y^2}{4}=61y-y^2 \\ &\qquad \left(x+\frac{y}{2}\right)^2=61y-\frac{3}{4}y^2 \\ &\qquad \left(2x+y\right)^2=244y-3y^2 \\ &\qquad -\frac{1}{3}\left(2x+y\right)^2=-\frac{122^2}{9}+\frac{122^2}{9}-\frac{244}{3}y+y^2 \\ &\qquad -\frac{1}{3}\left(2x+y\right)^2=-\frac{122^2}{9}+\left(\frac{122}{3}-y\right)^2 \\ &\qquad -3(2x+y)^2=-122^2+(122-3y)^2 \\ &\qquad (3y-122)^2+3(2x+y)^2=122^2=14884=2^2 \cdot 61^2 \implies\\ &\left[\alpha^2+3 \beta ^2=14884\right] \quad | \quad \alpha=3y-122 \ \ \land \ \ \beta= 2x+y \end{align}$$

Whatever $\alpha,\beta$ that might solve the above $$\begin{align}y=\frac{\alpha+122}{3} \ \ \land \ \ x&=\frac{\beta-y}{2} \\ &=\frac{\beta-\frac{\alpha+122}{3}}{2} \\ &=\frac{3\beta-\alpha -122}{6}\end{align}$$

$$ \underline{(\alpha_i, \beta_i) \to (x_i, y_i, z_i)} \\ (-74,56) \to (20,16,25) \\ (-74,-56) \to (-36,16,81) \\ (-47,65) \to (20,25,16) \\ (-47,-65) \to (-45, 25,81) \\ (121,9) \to (-36,81,16) \\ (121,-9) \to (-45,81,25) \\ $$

$(\alpha, \beta)$ Solutions obtained through the help of Darios online equation solver: https://www.alpertron.com.ar/QUAD.HTM

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