Are there nonzero natural numbers such that $\sqrt{4n+5}+\sqrt{5n+1}+\sqrt{9n+4}= \frac{nx}{y}$?

Question

Check if there are nonzero natural numbers $n,x,y$ such that:

$$\sqrt{4n+5}+\sqrt{5n+1}+\sqrt{9n+4}= \frac{nx}{y}. $$Thank you in advance!

My ideas

So we can simply show that $4n+5,5n+1,9n+4$ are perfect squares. But I don't know what to do forward, how to show how each equal. Hope one of you can help me!

Answer 1

Let me try to explain the concept of "mod". When we write $$ a\equiv b\pmod m $$ all that means is that $b-a$ is a multiple of $m$. Saying the same thing another way, it means that $a$ and $b$ leave the same remainder when you divide by $m$. Vocabulary: in this case, we say $a$ is congruent to $b$, modulo $m$.

Note that when you divide by $m$, the remainder will be one of the numbers $0,1,2,\dotsc,m-1$. So every $a$ is congruent to exactly one of the numbers $0,1,2,\dotsc,m-1$.

For example, every $x$ is congruent to exactly one of the numbers $0,1,2,\dotsc,8$ modulo $9$. Then, $x^2$ is congruent to exactly one of the numbers $0^2,1^2,2^2,\dotsc,8^2$ modulo $9$. Now those numbers are $0,1,4,9,16,25,36,49,64$, and when you divide each of these in turn by $9$, you get the remainders $0,1,4,0,7,7,0,4,1$. So we have established that the only possible remainders when you divide $x^2$ by $9$ are $0,1,4$ and $7$.

For $\sqrt{4n+5}$ to be an integer, $4n+5$ has to be a square, so we must have $$ 4n+5\equiv0,1,4{\rm\ or\ }7\pmod9 $$ Subtract $5$ from both sides to get $$ 4n\equiv4,5,8{\rm\ or\ }2\pmod9 $$ (How did I do that? $4n+5\equiv0\pmod9$ is $4n\equiv-5\pmod9$ but $-5\equiv4\pmod9$ since $4-(-5)=9$ is a multiple of $9$, so $4n+5\equiv0\pmod9$ becomes $4n\equiv4\pmod9$. Similarly for the other numbers.)

Now, if I multiply both sides by $7$, I get $$ 28n\equiv28,35,56{\rm\ or\ }14\pmod9 $$ and replacing all those numbers by their remainders on division by $9$ we get $$ n\equiv1,8,2{\rm\ or\ }5\pmod9 $$

Now, for $\sqrt{5n+1}$ to be an integer, $5n+1$ must be a square, so $$ 5n+1\equiv0,1,4{\rm\ or\ }7\pmod9 $$ By the same kind of calculations as done above, we get $$ 5n\equiv8,0,3{\rm\ or\ }6\pmod9 $$ and then multiplying by $2$ and taking remainders we get $$ n\equiv7,0,6{\rm\ or\ }3\pmod9 $$ But that is incompatible with the conditions we found for $4n+5$ to be a square. So there is no $n$ for which $4n+5$ and $5n+1$ are both squares, so there is no $n$ satisfying the original equation.

Answer 2

Assume that all three radicands are perfect squares.

The least common multiple of 4, 5, and 9 is 180 so let's do arithmetic modulo 180. Though brute force, I found that all perfect squares must be in the set $\{0, 1, 4, 9, 16, 25, 36, 40, 45, 49, 61, 64, 76, 81, 85, 100, 109, 121, 124, 136, 144, 145, 160, 169\}$ (mod 180).

So $4n+5$ to be a perfect square, we must have $n \in \{1, 5, 10, 11, 14, 19, 20, 26, 29, 35, 41, 44, 46, 50, 55, 56, 59, 64, 65, 71, 74, 80, 86, 89, 91, 95, 100, 101, 104, 109, 110, 116, 119, 125, 131, 134, 136, 140, 145, 146, 149, 154, 155, 161, 164, 170, 176, 179\}$ (mod 180).

For $5n + 1$ to be a perfect square, we must have $n \in \{0, 3, 7, 12, 15, 16, 24, 27, 36, 39, 43, 48, 51, 52, 60, 63, 72, 75, 79, 84, 87, 88, 96, 99, 108, 111, 115, 120, 123, 124, 132, 135, 144, 147, 151, 156, 159, 160, 168, 171\}$ (mod 180).

For $9n+4$ to be a perfect square, we must have $n \in \{0, 4, 5, 8, 9, 13, 20, 24, 25, 28, 29, 33, 40, 44, 45, 48, 49, 53, 60, 64, 65, 68, 69, 73, 80, 84, 85, 88, 89, 93, 100, 104, 105, 108, 109, 113, 120, 124, 125, 128, 129, 133, 140, 144, 145, 148, 149, 153, 160, 164, 165, 168, 169, 173 \}$ (mod 180).

For all three to be perfect squares, you need $n$ to be in all three of these sets simultaneously. However, it turns out that the intersection of the sets is empty, so no such $n$ can exist.

Answer 3

We can use the concept of digital root here, as explained in the link.

Let $dr(x)$ be defined as an iterative function that calculates the digital root of any natural number $x$. Now, using the statement for the rule of products of digital roots given in the link: $$dr(x^2)=dr(x*x)=dr(dr(x)*dr(x))=dr(dr(x)^2)$$

But $dr(x)$ can only be a natural number less than or equal to $9$. Putting in $dr(x)=1$ to $dr(x)=9$ on the RHS and calculating the value of the RHS, we find that the only possible values are the numbers in the set $\{1, 4, 9, 7\}$. Therefore, the only possible values of the digital roots of perfect squares of natural numbers are those in $\{1, 4, 9, 7\}$. If we can prove that the digital root of a particular number does not fall in this set, then it will imply that the given number is not a perfect square.

Now consider the three terms $4n+5$, $5n+1$ and $9n+4$. Using the product and addition rules, their digital roots are: $$dr(4n+5)=dr(4dr(n)+5)$$ $$dr(5n+1)=dr(5dr(n)+1)$$ $$dr(9n+4)=dr(9dr(n)+4)$$

Now we can put in values of $dr(n)$ from 1 to 9. For every such value of $dr(n)$, at least one of the three expressions on the RHS evaluates to a value that is not in $\{1, 4, 9, 7\}$. For example if the digital root of $n$ is 5, then the digital root of $5n+1$ is 8, which is not permissible for a perfect square, while for digital root of $n$ being 8, digital root of $5n+1$ is 5, which is also not permissible. Similarly we can check for all values of digital sum of $n$ from 1 to 9. [NOTE: For the 3rd term, its digital root is always 4, using the rule for casting out nines.] Therefore the 3 terms cannot be simultaneously perfect squares. So it is impossible for the sum of their square roots to be a rational number.

I also claim that the third term is redundant, since its digital sum is always 4, which is permissible for squares. Even if the last term is removed from the question, no value of $n$ can be found.

Answer 4

Put $4n+5=x^2$ so $5n+1=\dfrac{5x^2-21}{4}$ and $9n+4=\dfrac{9x^2-29}{4}$ and we need that $5x^2-21$ and $9x^2-29$ be squares. Since $$5x^2-21=y^2\Rightarrow 5x^2\equiv y^2\pmod7$$ and $5$ is not a square modulo $7$ only $x=y=0\pmod7$ but in this case $-29=-1\pmod7$ and $-1$ is not square modulo $7$. Thus there are not solution for this problem.

Answer 5

Mod $7$ seems to be most economic as Piquito tried.

$4n+5\equiv m^2 \pmod{7}\implies\{1,4,5,6\},$

$5n+1\equiv m^2 \pmod{7}\implies\{0,2,3,4\},$

$9n+4\equiv m^2 \pmod{7}\implies\{0,2,5,6\}$ modulo 7.

The intersection of sets is empty. So there is no solution. I used WolframAlpha to solve the modular arithmetic equations but they can be doable by hand since $7$ is small compared to $180$.