# Integer roots of $(6x^2+15xy)^{1/2}+y^2=10$ and $x+y=3$

I'm trying to figure out how to solve the following equations:

Find the integer roots of

$$(6x^2+15xy)^{1/2}+y^2=10 \quad\text{and}\quad x+y=3$$

I tried substituting $y=3-x$ to $(6x^2+15xy)^{1/2}+y^2=10$, but I ended up having more square roots and the power of $x$ just goes higher and higher.

Please help me with a very detailed answer, as I am trying to understand how to this step by step. Thank you very very much!

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Please click the "edit" button to see how I formatted your post. All that HTML crud you put in just wasted time because it all had to be erased. – dfeuer Oct 5 '13 at 22:57
@dfeuer thanks man. I'm a first time user – Confusedwithmath Oct 5 '13 at 23:02

Hint $y^2 \leq 10$

There are not too many integers $y$ which satisfy this inequality.....

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So how to do it? – Confusedwithmath Oct 5 '13 at 23:09
@Confusedwithmath Which are all the integers $y$ which satisfy $y^2 \leq 10$? – N. S. Oct 5 '13 at 23:11
1,2,3. I know that the answer is x=1 and y=2, because x+y =3 can only either be 0+3 or 1+2 or 2+1, but how do you get y^2≤10? – Confusedwithmath Oct 5 '13 at 23:19
@Confusedwithmath Those are the positive integers..... And each value of $y$ yields an $x$ such that $x+y=3$. Then you need to check if that pair satisfies the first equation... – N. S. Oct 5 '13 at 23:29
@Confusedwithmath And $y^2 \leq y^2+\sqrt{6x^2+15xy}=3$. – N. S. Oct 5 '13 at 23:30

By definition, both $\sqrt{6x^2+15xy}$ and $y^2$ are nonnegative. So the left-hand side of the equation is the sum of two nonnegative numbers, one of which is a square, whose sum is $10$. If $\lvert y \rvert \ge 4$, then $y^2 \ge 16$, and the equation can never be satisfied. Hence, you need only consider $y \in \{-3,-2,-1,0,1,2,3\}$.

Using the second equation, $x+y=3$, each of those possibilities for $y$ yields exactly one $x$, i.e. $x=3-y$. So let's start with $y=-3$. This gives $x=3-y=3-(-3)=6$. Now substituting $(x,y)=(6,-3)$ into the first equation implies \begin{align} 10 &= \sqrt{6(6)^2+15(6)(-3)} + (-3)^2 \\ &= \sqrt{-54} + 9 \\ 1 &= \sqrt{-54}, \end{align} which is obviously a contradiction.

Hope this helps!

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