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

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.

  • $\begingroup$ Thanks, Took a little time but this is much better.... $\endgroup$ – Shh Apr 6 '18 at 7:03

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 $y=16, z=25, x=20$

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.