Adithya from Hymers College in the UK used a method similar to Charlie's method:

For starters, I approximated the values of the two surds to give the two rational numbers 8.06 and 8.19 (both to two decimal places). Then I thought it would be most suitable if I listed all the different fraction denominators from 1-25 in a table. Initially, it seemed to me that 25 would be a logical and sensible stopping point in the sense that fractions with a denominator higher than 25 have smaller decimal values and therefore are more likely to have more than one apparent fraction between 8.06 and 8.19. For certain denominators, for example a fraction whose denominator is 13, I have only listed the relevant decimal values as there are numerous. Those fractions such as 2/16 haven’t been listed as another decimal value due to the fact that they cancel down.

So in the 13 row of the table, only .0, .076923.. and .15384... are shown, since some have been found that are between .06 and .19 (the decimal parts of $\sqrt{75}$ and $\sqrt{67}$), but the next one would be .23077 which is more than .19. In the 16 row of the table, .125 is not shown as it already occurs in the 8 row.

Kate, Noah and Anabella from Our Lady of Lourdes School in Canada began using Alison's method, and then proved that they had found all of the possible denominators. This is their work:

If $q=1:$ $65\times1 < p^2<67\times1\Rightarrow65< p^2<67$

$8^2 = 64,$ $9^2 = 81.$

There are no fractions with denominators of 1 between the two radicals.

For starters, I approximated the values of the two surds to give the two rational numbers 8.06 and 8.19 (both to two decimal places). Then I thought it would be most suitable if I listed all the different fraction denominators from 1-25 in a table. Initially, it seemed to me that 25 would be a logical and sensible stopping point in the sense that fractions with a denominator higher than 25 have smaller decimal values and therefore are more likely to have more than one apparent fraction between 8.06 and 8.19. For certain denominators, for example a fraction whose denominator is 13, I have only listed the relevant decimal values as there are numerous. Those fractions such as 2/16 haven’t been listed as another decimal value due to the fact that they cancel down.

So in the 13 row of the table, only .0, .076923.. and .15384... are shown, since some have been found that are between .06 and .19 (the decimal parts of $\sqrt{75}$ and $\sqrt{67}$), but the next one would be .23077 which is more than .19. In the 16 row of the table, .125 is not shown as it already occurs in the 8 row.

terminating fraction denominators | recurring fraction denominators |

2: .0 or .5 | |

3: .0 or .3... or .6... | |

4: .0 or .25 or .75 | |

5: .0 or .2 or .4 or .6 or .8 | |

6: .0 or .16... | |

7: .0 or .14285... | |

8: .0 or .125 | |

9: .0 or .1... | |

10: .0 or .1 | |

11: .0 or .0909... or .1818... | |

12: .0 or .0833... | |

13: .0 or .07692 or .15384... | |

14: .0 or .07142... | |

15: .0 or .0666... or .1333... | |

16: .0 or .0625 or .1875 | |

17: .0 or .05882... or .11764... or .17647 | |

18: .0 or .0555... | |

19: .0 or .05263... or .10526... or .15789... | |

20: .0 or .05 or .15 | |

21: .0 or .04761... or .09523... | |

22: .0 or .04545... or .13636... | |

23: .0 or .043478... or .08695... or .13043... or .17391 | |

24: .0 or .041666... | |

25: .0 or .04 or .08 or .12 or .16 |

If we express the difference between the fractions $\dfrac{p}{q}$ and $\dfrac{p+1}{q}$ as $\dfrac{1}{q}$, then as the number $q$ starts to increase the value of $\dfrac{1}{q}$ starts to decrease due to the fact that they are inversely proportional. This means to say that the larger the number $q$ is the more likely there is to be a fraction between the interval 8.06 and 8.19 compared to a
smaller number $q$. As I’ve clearly shown in the table above, all denominators barring 2,3,4 and 5 have a fraction between the approximate numbers of 8.06 and 8.19 which represent $\sqrt{65}$ and $\sqrt{67}$. In conclusion, there are no rational numbers in the interval between $\sqrt{65}$ and $\sqrt{67}$ with denominators of 1,2,3,4 and 5.

Gabriel from Wheaton North High School in the USA also used a table. This is some of Gabriel's work:

Knowing that the fraction that I used had to be slightly greater than eight, I {used the expression} $\dfrac{8n+1}{n}.$

Number ($n$) | $\frac{8n+1}{n}$ | Parameter satisfaction? |

1 | 9.00 | No |

2 | 8.50 | No |

3 | 8.33 | No |

4 | 8.25 | No |

5 | 8.20 | No |

6 | 8.1666667 | Yes |

7 | 8.1428571 | Yes |

8 | 8.1250000 | Yes |

9 | 8.1111111 | Yes |

10 | 8.1000000 | Yes |

11 | 8.0909091 | Yes |

12 | 8.0833333 | Yes |

13 | 8.0769231 | Yes |

14 | 8.0714286 | Yes |

15 | 8.0666667 | Yes |

16 | 8.0625000 | Yes |

17 | 8.0588235 | No |

18 | 8.0555556 | No |

19 | 8.0526136 | No |

20 | 8.0500000 | No |

Then Gabriel realised that this table only included fractions $8\frac{1}{n}$ and ignored fractions with numerators greater than $1$. Considering the $17$ row, Gabriel said:

$\dfrac{x}{17}$ will always be a number with a decimal of $.058824,$ $.117647,$ $0.176471,$ and so on.

Kate, Noah and Anabella from Our Lady of Lourdes School in Canada began using Alison's method, and then proved that they had found all of the possible denominators. This is their work:

If $q=1:$ $65\times1 < p^2<67\times1\Rightarrow65< p^2<67$

$8^2 = 64,$ $9^2 = 81.$

There are no fractions with denominators of 1 between the two radicals.

If $q=2:$ $65\times4< p^2 <67\times4\Rightarrow260 < p^2 <268$

$16^2 = 256,$ $17^2 = 289$

There are no fractions with denominators of 2 between the two radicals

If $q=3:$ $65\times9< p^2 <67\times9\Rightarrow585< p^2 <603$

$24^2 = 576,$ $25^2 = 625$

There are no fractions with denominators of 3 between the two radicals

If $q=4:$ $65\times16< p^2 <67\times16\Rightarrow1040 < p^2 <1072$

$32^2 = 1024,$ $33^2 = 1089$

There are no fractions with denominators of 4 between the two radicals

If $q=5:$ $65\times25 < p^2 <67\times25\Rightarrow1625 < p^2 <1675$

$40^2 = 1600,$ $41^2 = 1681.$

There are no fractions with denominators of 5 between the two radicals

If $q=6:$ $65\times36 < p^2 <67\times36\Rightarrow2340 < p^2 <2412$

$48^2 = 2304,$ $49^2 = 2401,$ $50^2 = 2500$

There is 1 fraction ($\frac{49}{6}$) with denominator of 6 between the two radicals

If $q=7:$ $65\times49 < p^2 <67\times49\Rightarrow3185 < p^2 <3283$

$56^2 = 3136,$ $57^2 = 3249,$ $58^2 = 3364$

There is 1 fraction ($\frac{57}{7}$) with denominator of 7 between the two radicals

If $q=8:$ $65\times64 < p^2 <67\times64\Rightarrow4160 < p^2 <4288$

$64^2 = 4096,$ $65^2 = 4225,$ $66^2 = 4356$

There is 1 fraction ($\frac{65}{8}$) with denominator of 8 between the two radicals

If $q=9:$ $65\times81 < p^2 <67\times81\Rightarrow5262 < p^2 <5427$

$72^2 = 5184,$ $73^2 = 5329,$ $74^2 = 5476$

There is 1 fraction with denominator of 9 between the two radicals

If $q=10:$ $65\times100 < p^2 <67\times100\Rightarrow6500 < p^2 <6700$

$80^2 = 6400,$ $81^2 = 6561,$ $82^2 = 6724$

There is 1 fraction with denominator of 10 between the two radicals

If $q=11:$ $65\times121 < p^2 <67\times121\Rightarrow9360 < p^2 <9648$

$88^2 = 7744,$ $89^2 = 7921,$ $90^2 = 8100,$ $91^2 = 8281$

There are 2 fractions with denominators of 11 between the two radicals

Now that we have proven that all integer denominators from 1-5 have no numerator which places them in between the two radicals, 6-10 have one, and 11 has two, we are going to prove that there can be no fractions with a integer denominator greater than 11 that do not have a numerator that places them in between the two radicals.

Here is a hypothetical proof involving variables which will be applied to the problem.

Set variables $x,$ $y,$ $\dfrac{a}{z}$ and $\dfrac{b}{z}$ where $x<\dfrac{a}{z}<\dfrac{b}{z}$ and $a+1=b.$ Since $\dfrac{a}{z}$ and $\dfrac{b}{z}$ share the common denominator $z,$ $\dfrac{b}{z}-\dfrac{a}{z}=\dfrac{1}{z}$ and $\dfrac{1}{z} < y-x.$

Since when you increase the denominator but leave the numerator unchanged the fraction decreases in value, we can say then that $\dfrac{1}{z+c} < y-x$ {whenever} $c > 0.$

Now we will substitute the variables for the problem’s numbers to prove that every denominator greater than 11 has a numerator that would place it in between the radicals:

$$\sqrt{65}<\dfrac{89}{11}<\dfrac{90}{11}<\sqrt{67}$$

Since $89+1=90$ and the two fractions share a common denominator, $\dfrac{90}{11}-\dfrac{89}{11}=\dfrac{1}{11}$ and $\dfrac{1}{11}<\sqrt{67}-\sqrt{65}$

Since when you increase the denominator but leave the numerator unchanged the fraction decreases in value we can say then that $\dfrac{1}{11+c}<\sqrt{67}-\sqrt{65}$ {for any} $c>0.$

Since any fraction with 1 as a numerator and a number greater than 11 as a denominator is less than the distance between the two radicals at least one multiple of that fraction must land in between $\sqrt{65}$ and $\sqrt{67}$.

Therefore the above denominators below 11 which do not have a numerator that places them in between $\sqrt{65}$ and $\sqrt{67}$ must be the only denominators that conform to that rule.

Q.E.D.

Amrit from Hymers College used a purely algebraic approach:

We want to find fractions in the form of $\dfrac{p}{q}$ such that $$\sqrt{65} < \frac{p}{q} < \sqrt{67}$$

Squaring all parts of inequality and rearranging, we have $$65(q^2) < p^2 < 67(q^2)$$

If this range lies between two consecutive square numbers, $p$ cannot exist (we are dealing with integers for p and q). For example, if $65(q^2) > 16^2$ but $67(q^2) < 17^2,$ there is no integer $p$ which lies in between $16$ and $17$.

Therefore, $p$ cannot exist if there is a number $x$ such that $$x^2 < 65(q^2) < p^2 < 67(q^2) < (x+1)^2$$

And we will always be able to find a square between two numbers that do not lie between two consecutive squares, so if there is not such a number $x$, then $p$ does exist.

Breaking up the inequality, we have $x^2 < 65(q^2)$ and $(x+1)^2 > 67(q^2).$

Square rooting both sides of both inequalities and making the subject $x,$ we have $x<\sqrt{65}\times q$ or $x > \sqrt{67}\times q-1$.

So if there is an integer $x$ that lies between these two bounds, then $p$ cannot exist, and if there is not, then $p$ does exist.

The inequality suggests that $\sqrt{67}\times q-1 < \sqrt{65}\times q,$ but there is a $q$ such that $\sqrt{67}\times q-1 > \sqrt{65}\times q$

Subtracting $\sqrt{65}\times q$ and adding $1$ to both sides, we get $\sqrt{67}q-\sqrt{65}q > 1$

Grouping in terms of $q$, we get $\left(\sqrt{67}-\sqrt{65}\right)q > 1$

Dividing both sides by $\sqrt{67}-\sqrt{65}$, we get $q > \dfrac{1}{\sqrt{67}-\sqrt{65}}$ which is approximately equal to $8.12,$ so $q > 8.12$.

This constraint applies for the first integer after $8.12,$ which is $9.$ Therefore, when $q \ge 9,$ $\sqrt{67}\times q - 1 > \sqrt{65}\times q$, so it is impossible to find an $x$ that is larger than $\sqrt{67}\times q-1 $ but smaller than $\sqrt{65}\times q$ - so there will be a $p$ such that $\sqrt{65}<\dfrac{p}{q}<\sqrt{67}.$

So fractions exist with a denominator of $9$ or greater that lie between $\sqrt{65}$ and $\sqrt{67}.$