1. Find all integers $a, b, c$ and $d$ that satisfy the equation

$$
a \sqrt{2}+b \sqrt{5}+c=d \sqrt{10}
$$
IV /1. Clearly, one solution is $a=b=c=d=0$. We will show that this is the only one. Suppose there is another solution $(a, b, c, d)$. We can assume that the numbers $a, b, c$, and $d$ are coprime, otherwise, the greatest common divisor could be canceled out from the equation (since not all numbers are zero, the greatest common divisor exists). Rearrange the equation to $a \sqrt{2}+b \sqrt{5}=d \sqrt{10}-c$ and square it to get

$$
2 a^{2}+5 b^{2}-c^{2}-10 d^{2}=2 \sqrt{10}(c d-a b)
$$

Since $\sqrt{10}$ is not a rational number, we must have $2 a^{2}+5 b^{2}-c^{2}-10 d^{2}=0$ or

$$
2 a^{2}-c^{2}=10 d^{2}-5 b^{2}
$$

Since the right-hand side is divisible by 5, the left-hand side must also be. However, the square of a natural number gives a remainder of 0, 1, or 4 when divided by 5. Therefore, the only possibility is that the remainders of $a^{2}$ and $c^{2}$ when divided by 5 are both 0, which means that $a$ and $c$ are divisible by 5. Thus, the left-hand side is divisible by 25, and so must the right-hand side. This means that $2 d^{2}-b^{2}$ must be divisible by 5. Similarly as before, we conclude that $b$ and $d$ are divisible by 5. Therefore, all numbers $a, b, c$, and $d$ are divisible by 5, which is a contradiction since they are coprime.

![](https://cdn.mathpix.com/cropped/2024_06_07_550bffe3dbbdd9dbd37bg-13.jpg?height=54&width=1633&top_left_y=687&top_left_x=220)

Reformulating and appropriately squaring the equation (at most one root remains) .. 2 points

![](https://cdn.mathpix.com/cropped/2024_06_07_550bffe3dbbdd9dbd37bg-13.jpg?height=51&width=1633&top_left_y=791&top_left_x=220)

Realizing that divisibility by 5 leads to a contradiction ............................. 3 points

2. method. If $(a, b, c, d)$ is a solution to the equation, as in the first solution, we get that $2 a^{2}+5 b^{2}-c^{2}-10 d^{2}=0$ must hold. If we square the equation in the form $a \sqrt{2}+c=d \sqrt{10}-b \sqrt{5}$, with similar reasoning, we get that $2 a^{2}+c^{2}-10 d^{2}-5 b^{2}=0$ must hold. If we subtract these two equations, we get $10 b^{2}-2 c^{2}=0$ or $c^{2}=5 b^{2}$. Therefore, $c= \pm \sqrt{5} b$, but since $\sqrt{5}$ is not a rational number, it follows that $b=c=0$. From one of the equations, we then get $a^{2}=5 d^{2}$, from which we get $a=d=0$. Thus, the equation has only one solution $a=b=c=d=0$.

![](https://cdn.mathpix.com/cropped/2024_06_07_550bffe3dbbdd9dbd37bg-13.jpg?height=52&width=1633&top_left_y=1256&top_left_x=220)

Reformulating and appropriately squaring the equation (at most one root remains) .. 2 points

![](https://cdn.mathpix.com/cropped/2024_06_07_550bffe3dbbdd9dbd37bg-13.jpg?height=54&width=1633&top_left_y=1355&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_550bffe3dbbdd9dbd37bg-13.jpg?height=55&width=1633&top_left_y=1406&top_left_x=220)

3. method. Squaring the equation $a \sqrt{2}+b \sqrt{5}=d \sqrt{10}-c$, we get $2 a^{2}+5 b^{2}-c^{2}-10 d^{2}=$ $(-a b-c d) \sqrt{10}$. It must hold that $a b=-c d$. If $d=0$, then $a b=0$ and we get $a \sqrt{2}+c=0$ or $b \sqrt{5}+c=0$. In both cases, we get the solution $a=b=c=d=0$. Suppose $d \neq 0$, then $c=-\frac{a b}{d}$. Substituting into the original equation, we get $a d \sqrt{2}+b d \sqrt{5}-a b-d^{2} \sqrt{10}=0$, or $(d \sqrt{5}-a)(b-d \sqrt{2})=0$. Since $d \neq 0$, we do not get a new solution.

![](https://cdn.mathpix.com/cropped/2024_06_07_550bffe3dbbdd9dbd37bg-13.jpg?height=278&width=1654&top_left_y=1777&top_left_x=215)