# Diophantine equation solving method

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?

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)

• oops, I solved the wrong exercise. I think it is now correct – pendermath May 2 at 12:33

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.

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