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So, the equation is $x(x-4)+y(6-y)=10$. Now, I was thinking of checking all pairs of $z+t=10$ if they can be substituted in the equation and get an integer result. Is there a more efficient method?

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Working out, you get $$ (x-2)^2 -(y-3)^2 = 5 \Rightarrow (x+y-5)(x-y+1)=5 $$ As 5 is prime there are only two possibilities, which lead to the two solutions: (5,1) and (5,5)

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  • $\begingroup$ oops, I solved the wrong exercise. I think it is now correct $\endgroup$ – pendermath May 2 at 12:33
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We have $$x^2 -4x +(6y-y^2-10)=0$$ $$\implies x= \frac{4\pm \sqrt{4y^2-24y+56}}{2}=2\pm\sqrt{y^2-6y+14}\\ = 2\pm\sqrt{(y-3)^2 + 5}$$

For $x$ to be an integer, the term inside the radical should be a perfect square. Notice that the only pair of squares that are $5$ apart is $(4,9)$, as the difference between consecutive squares increases as we go further.

Therefore, $$(y-3)^2 = 4 \implies y=3\pm 2 = 5,1$$

and $$x=2\pm 3=5 \ \text{(neglecting the negative solution)} $$ and we’re done.

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  • $\begingroup$ Observe that x cannot be negative, as this is a Diophantine equation $\endgroup$ – pendermath May 2 at 12:32
  • $\begingroup$ @pendermath Corrected. $\endgroup$ – Tavish May 2 at 12:59

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