Not 1?

Suppose $\tfrac{x^2+3}{247}$ is an integer. Then, $x^2 \equiv -3 \pmod{247}$ .

Since $247 = 13 \cdot 19$ , by the Chinese Remainder Theorem, $x^2 \equiv -3 \equiv 10 \pmod{13}$ and $x^2 \equiv -3 \equiv 16 \pmod{19}$ .

In $(\text{mod} \ 13)$ , we have $0^2 \equiv 0$ , $(\pm 1)^2 \equiv 1$ , $(\pm 2)^2 \equiv 4$ , $(\pm 3)^2 \equiv 9$ , $(\pm 4)^2 \equiv 3$ , $(\pm 5)^2 \equiv 12$ , $(\pm 6)^2 \equiv 10$ .

In $(\text{mod} \ 19)$ , we have $0^2 \equiv 0$ , $(\pm 1)^2 \equiv 1$ , $(\pm 2)^2 \equiv 4$ , $(\pm 3)^2 \equiv 9$ , $(\pm 4)^2 \equiv 16$ , $(\pm 5)^2 \equiv 6$ , $(\pm 6)^2 \equiv 17$ , $(\pm 7)^2 \equiv 11$ , $(\pm 8)^2 \equiv 7$ , $(\pm 9)^2 \equiv 5$ .

Therefore, we must have $x \equiv \pm 6 \pmod{13}$ and $x \equiv \pm 4 \pmod{19}$ .

Using the Chinese Remainder Theorem, we find the following: $x \equiv -6 \pmod{13}$ and $x \equiv -4 \pmod{19}$ $\leadsto x \equiv \ \ 72 \pmod{247}$ $x \equiv +6 \pmod{13}$ and $x \equiv -4 \pmod{19}$ $\leadsto x \equiv 110 \pmod{247}$ $x \equiv -6 \pmod{13}$ and $x \equiv +4 \pmod{19}$ $\leadsto x \equiv 137 \pmod{247}$ $x \equiv +6 \pmod{13}$ and $x \equiv +4 \pmod{19}$ $\leadsto x \equiv 175 \pmod{247}$

Therefore, we must have $x \equiv 72$ , $110$ , $137$ , $175 \pmod{247}$ .

Since $\tfrac{x^2+3}{247}$ is positive and increasing for $x > 0$ , the smallest positive integer value of $\tfrac{x^2+3}{247}$ is attained by the smallest positive integer $x$ such that $\tfrac{x^2+3}{247}$ is an integer, i.e. $x = 72$ . This gives $\tfrac{x^2+3}{247} = \tfrac{72^2+3}{247} = \boxed{21}$ .

We only need to find the smallest integer that $\begin{aligned} x^2\equiv& 10\mod{13}\\ x^2\equiv& 16\mod{19} \end{aligned}$ Refer to Quadratic Residues - Primes , we know that

$\begin{aligned} x=&13k+6\mbox{ or }x=13k+7\\ x=&19k+4\mbox{ or }x=19k+15 \end{aligned}$

Solving them by the Chinese Remainder Theorem , we have $x=72,110,137,175$ . So $x=72$ , which gives $(x^2+3)/247=21$ .

Note that $247 = 13 \cdot 19$ , hence we need $13 \mid x^2 + 3$ and $19 \mid x^2 + 3$ , i.e. $\begin{cases} x^2 \equiv 10 \pmod{13} \\ x^2 \equiv 16 \pmod{19} \end{cases}$ Examining quadratic residues modulos $13,19$ we find that it's equivalent to $\begin{cases} x \equiv \langle 6,7 \rangle \pmod{13} \\ x \equiv \langle 4,15 \rangle \pmod{19} \end{cases}$ where the notation $\langle a,b \rangle$ denotes exactly one of $a,b$ (this is not standard notation). Hence there exists $a \in \mathbb{N} \cup \{0\}$ such that $x = 19a + \langle 4,15 \rangle$ So in first congruence $6a \equiv \langle 6,7 \rangle - \langle 4,15\rangle \pmod{13}$ i.e. $a \equiv \frac{\langle 6,7 \rangle - \langle 4,15\rangle}{6} \equiv (-2) \cdot \big(\langle 6,7 \rangle - \langle 4,15\rangle\big) \pmod{13}$ where the fraction doesn't denote actual division by $6$ , but modular multiplicative inverse of $6$ in modulo $13$ .

1) $\langle 4,15 \rangle = 4$ and $\langle 6,7 \rangle = 6$ , then $a \equiv (-2) \cdot 2 \equiv 9 \pmod{13}$ . So there exists $b \in \mathbb{N} \cup \{0\}$ such that $a = 13b + 9$ , i.e. $x = 247b + 175$ .

2) $\langle 4,15 \rangle = 15$ and $\langle 6,7 \rangle = 6$ , then $a \equiv (-2) \cdot (-9) \equiv 5 \pmod{13}$ . So there exists $b \in \mathbb{N} \cup \{0\}$ such that $a = 13b + 5$ , i.e. $x = 247b + 110$ .

3) $\langle 4,15 \rangle = 4$ and $\langle 6,7 \rangle = 7$ , then $a \equiv (-2) \cdot 3 \equiv 7 \pmod{13}$ . So there exists $b \in \mathbb{N} \cup \{0\}$ such that $a = 13b + 7$ , i.e. $x = 247b + 137$ .

4) $\langle 4,15 \rangle = 15$ and $\langle 6,7 \rangle = 7$ , then $a \equiv (-2) \cdot (-8) \equiv 3 \pmod{13}$ . So there exists $b \in \mathbb{N} \cup \{0\}$ such that $a = 13b + 3$ , i.e. $x = 247b + 72$ .

Hence all $x \in \mathbb{N}$ such that $247 \mid x^2 + 3$ are $x \in \Big\{247b + c \mid b \in \mathbb{N} \cup \{0\} \wedge c \in \{72,110,137,175\}\Big\}$ It's obvious that smallest $\frac{x^2 + 3}{247}$ will have smallest $x$ and smallest $x$ is $72$ , hence the value sought is $\frac{72^2 + 3}{247} = \boxed{21}$

I whipped up a quick little C program to help me here:

int i = 0, num = 1.5, answer = 1;

do { num = ( i * i + 3 );
i += 1;
answer = num/247;
}
while ( num % 247 != 0 );

printf("%i\n", answer);

$\frac{x^2+3}{247}=k\Leftrightarrow x^2+3=13\cdot 19k$

$\Rightarrow$ when $x^2$ is divided by $13$ , the remainder is $10$

$\Rightarrow \left[\begin{matrix}x=13a+6 \\ x=13a+7\end{matrix},\right.$

Similarly, we also have $\Rightarrow \left[\begin{matrix}x=19b+4 \\ x=19b+15\end{matrix}.\right.$

Case 1: $x=13a+6=19b+4$

$\Rightarrow a=\frac{19b-2}{13}=b+\frac{6b-2}{13}$ ,

set $p=\frac{6b-2}{13}\Rightarrow b=\frac{13p+2}{6}=2p+\frac{p+2}{6}$ ,

set $t=\frac{p+2}{6}\Rightarrow p=6t-2$ ,

hence $b=13t-4,a=19t-6,x=247t-72$ ,

therefore, the smallest of $x^2$ in this case is $175^2.$

Similarly, in $3$ remaining cases, we have:

Case 2: $x=13a+6=19b+15$

$x=247t+357\Rightarrow x^2_{min}=110^2.$

Case 3: $x=13a+7=19b+4$

$x=247t-110\Rightarrow x^2_{min}=137^2.$

Case 4: $x=13a+7=19b+15$

$x=247t+319\Rightarrow x^2_{min}=72^2.$

The positive integer value of $k$ is the smallest if and only if the positive integer value of $x^2$ .

Besides, basing on $4$ above cases we have the smallest positive integer value of $x^2$ is $72^2$ ,

so $k_{min}=\frac{72^2+3}{247}=21.$

Since $247 = 13 \cdot 19,$ the problem is equivalent to solve the congruence equations: $x^2 \equiv -3 \pmod{13} \quad \text{and} \quad x^2 \equiv -3 \pmod{19}.$ By Lagrange's theorem, each of the congruence has at most two solutions. Also note that if $x_0$ is a solution to the congruence $x^2\equiv a \pmod{p}$ , then $p-x_0$ is also a solution to the congruence. In our case we can check very easily that $x\equiv 6, 7 \pmod{13}$ for the first congruence and $x\equiv 4,15 \pmod{19}$ for the second congruence. We then solve the four systems of linear congruence equations:

$\textbf{case I}$ $x\equiv 6 \pmod{13}$ and $x\equiv 4 \pmod{19}.$ The first congruence yields $x = 13k+6$ for some integer $k$ . Substituting into the second congruence, we have $13k \equiv -2 \pmod{19}.$ Multiplying by $3$ on both sides yields $k \equiv -6 \equiv 13 \pmod{19}.$ This implies $k =19 \ell + 13$ for some integer $\ell.$ Hence $x =13(19\ell+13)+6 = 247\ell + 175 \quad \text{or} \quad x\equiv 175 \pmod{247}.$

$\textbf{case II}$ $x\equiv 6 \pmod{13}$ and $x\equiv 15 \pmod{19}.$ By a similar argument as in case I, we get $x\equiv 110 \pmod{247}.$

$\textbf{case III}$ $x\equiv 7 \pmod{13}$ and $x\equiv 4 \pmod{19}.$ By a similar argument as in case I, we get $x\equiv 137 \pmod{247}.$

$\textbf{case IV}$ $x\equiv 7 \pmod{13}$ and $x\equiv 15 \pmod{19}.$ By a similar argument as in case I, we get $x\equiv 72 \pmod{247}.$

We conclude that the smallest positive integer $x$ such that $(x^2+3)/247$ is an integer is $x=72.$ This value of $x$ also yields the smallest integral value of $(x^2+3)/247$ since $x^2+3$ is positive. This is $(72^2+3)/247 =\boxed{ 21}.$

x^{2}=247k-3 x^{2}-16=13 19 k-19 (x+4)(x-4)=19(13 k-1) As 19 is a prime so, 19|x+4 or 19|x-4 x=19m-4 or x=19m+4 as output should be minimum putting x=19m-4 we get 361 m^{2}-152 m +19/247 =19(19 m^{2}-8 m+1)/13 19 =19 m^{2}-8 m+1/13 13m^{2}/13 -13*m/13 +(6m^{2}+5m+1)/13 after factorising we get (2m+1)(3m+1)/13 minimum of m is =4 so, x=72 and answwr is 21

The smallest value will come when x=72 So, by putting x=72 in the equation, the answer becomes 21

i wrote all the multiples of 247. The multiple should be in such a way on subtracting 3 from it, it should be a perfect square number 247*21=5187. 5187-3=5184 whose square root is 72(positive integer)

Make equation as [(247*x)-3]^(1/2) solution to this must be integer

put x=72,,, so we get 21 as the least positve integer for the above equation .

The smallest value will come when

x=72

So, by putting x=72 in the equation,

the answer becomes 21