Solving a Diophantine Fraction

Use long division, we can analyse the fraction:

$\frac {N^2-75}{5N+56} = \frac {(5N+56)(5N-56)+1261}{25(5N+56)} = \frac {5N-56+\frac {1261}{(5N+56)}}{25}$

Thus, $5N+56$ must be a divisor of $1261$ , so it's an element of $\{-1261, -97, -13, -1, 1, 13, 97, 1261\}$ . So the value of $N$ must be $241$ (other cases cause $N$ negative or not natural), and with $N = 241$ , the fraction is an integer.

Therefore, the solution of the problem is $241$ .

Note that $\dfrac{N^2-75}{5N+56}=\dfrac{1}{25}\cdot\dfrac{25N^2-1875}{5N+56}=\dfrac{1}{25}\left[5N-56+\dfrac{1261}{5N+56}\right]$ . It follows that $\dfrac{1261}{5N+56}$ is an integer, so $5N+56$ is a factor of $1261$ . Since the positive factors of $1261=13\cdot 97$ are $1, 13, 97, 1261$ , $N>0$ and $5N+56\equiv 1\pmod{5}$ , we get that $5N+56=1261$ or $N=241$ .

we assume the expression to be $k$ , then we have $N=\frac{5k\pm \sqrt{300+224k+25k^2}}{2}$

since N must integer so $300+224k+25k^2$ must be perfect square. so $300+224k+25k^2=L^2\Rightarrow (25k+112)^2-(5L)^2=5044$

note that when $N≤8$ there is no such $N$ is suitable . so $k$ must be $＞0$ .

now we divide the $5044=a\times b$ and assume $a>b$ ,then we have $a+b=2*(25k+112)>0,a-b=2*(5L)>0$ . $a-b$ must be multiple of $10$ . so $a=2\times 13\times 97,b=2,k=46,L=252$ .

finally, $N=\frac{5k\pm L}{2}=241,or,-11$ and the answer is $241$

given (5N+56)|(N^2-75) , then (5N+56)| (N(5N+56)-5(N^2-75)). this gives that (5N+56)| (56N+375) then 5N+56| (56(5N+56)-5(56N+375)) this gives that 5N+56|1261. but 1261=13*97. integer value of N is possible only if 5N+56=1261 this gives us n=241

The $N$ satisfies $N^2\equiv 75 \pmod{5N+56}$ Multiplying by $5^2$ yields $(5N)^2\equiv 25\cdot 75 \pmod{5N+56}$ which is the same as $(-56)^2\equiv 25\cdot 75 \pmod{5N+56}$ Hence, $5N+56 \mid (-56)^2-25\cdot 75=1261$ The divisors of $1261$ are $\{1,13,97,1261\}$ and only $1261$ is of the form $5N+56$ for some positive integer $N$ , namely $241$ .

by long division, $\frac{N^2-75}{5N+76}=\frac{1} {25}(5N-56+\frac{1261} {5N+56} )$ . Note that $1261=13⋅97$ .Then the only factor in the form of $5N+56$ is $1261$ , so $N=241$ .Indeed, as it gives $\frac{N^2-75}{5N+76}=46$ .So the only answer is $241$ .

$\frac{N ^ 2 - 75}{5N + 56}$ is an integer if and only if $5N + 56 \mid N ^ 2 - 75$ . This is equivalent to $5N + 56 \mid 5N ^ 2 - 375$ . Combined with the fact that $5N + 56 \mid 5N ^ 2 + 56N$ , we have $5N + 56 \mid 56N + 375 \Rightarrow 5N + 45 \mid 280N + 1875$ . Also $5N + 56 \mid 280N + 3136$ , so $5N + 56 \mid 1261$ . Since 1261 is prime, $N = \frac{1261 - 56}{5} = 241$ .

If the given fraction were to be an integer, so is any of its integral multiples. In particular if N^2 – 75 is multiplied by 25, it facilitates division by 5N + 56 as follows: First observe that (5N + 56) (5N – 56) = 25N^2 – 3136. So we write 25(N^2 – 75) = 25N^2 – 1875 = (25N^2 – 3136) + 1261. 25(N^2 – 75) / (5N + 56) = (5N – 56) + [1261 / (5N + 56)]. So it suffices to find N such that 1261 / (5N + 56) = 1. It is easy to see that N = 241 serves the purpose. By actual verification, with N = 241, we find the value of the given fraction = 58006/1261 = 46.

By long division, N^2 -75 = (5N+ 56)(N/5 - 56/25) + 1261/25. Multiplying throughout gives us 25(N^2 -75) = (5N+ 56)(5N-56) + 1261.

Either (5N +56) or (5N-56) is a factor of 1261. For N to be a positive integer < 1000, the only possibility is 5N + 56 = 1261 which gives us 5N = 1205 and hence N = 241. This yields (241^2-75) / (5 x 241+ 56) = 46.

Since $25(N^2-75) = (5N-56)(5N+56)+1261$ , it follows that $5N+56$ divides $N^2-75$ if and only if $5N+56$ divides $1261.$ Since $1261=13\cdot 97$ , the only possibilities are $5N+56$ is $1, 13, 97,$ or $1261$ . Since $N$ is required to be a positive integer no greater than $1000$ it follows that $5N+56 = 1261 \implies N=241$ .

We have to find an $N \in \mathbb{N}_{<1000}$ such that $\frac{N^2-75}{5N+56} \in \mathbb{Z}.$ So, $5N + 56 \mid N^2 - 75.$ It follows that $5N + 56 \mid N \cdot (5N + 56) - 5 \cdot (N^2 - 75) = 56N + 375.$ Similarly, $5N + 56 \mid (56N + 375) - 11 \cdot (5N + 56) = N - 241.$ But we know that for all $a \in \mathbb{Z}$ , $a \mid 0$ , so we just let $N-241 = 0 \implies N = 241$ . Let's try it out:

$\frac{N^2 - 75}{5N+56} = \frac{241^2-75}{5 \cdot 241 + 56} = \frac{58006}{1261} = 46 \in \mathbb{Z}.$ Hence, $N = \boxed{241}$ .

$\frac{N^{2} - 75}{5N + 56} = \frac{1}{25}(5N - 56 + \frac{1261}{5N + 56})$ . In order for the fraction to be an integer, $(5N + 56) \mid 1261$ . Since $1261 = 13 \times 97$ , and $N$ is a positive integer, the only possible value for $N$ is $241$ .  It turns out that $N = 241$ is indeed a solution. So, $N = 241$ .

$(5N+56)(5N-56) = 25N^2-3136.$

If $\frac{N^2-75}{5N+56}$ is an integer, then so is $\frac{25(N^2-75)}{5N+56} = \frac{25N^2 - 1875}{5N+56} = \frac{25N^2-3136 + 1261}{5N+56} = 5N-56 + \frac{1261}{5N+56}$ . For the latter to be an integer, $(5N+56) | 1261 = 13 \times 97$ . The only divisors of $1261$ which are congruent to $56 \mod 5$ are $1$ and $1261$ . If $5N+56=1$ then $N=-11<0$ . So, $5N+56 = 1261$ is the only possibility, and $N=241, \frac{N^2-75}{5N+56}=46$ .

Let $\frac{N^{2}-75}{5N+56}=k$ for some integer k. $N^{2}-75=5Nk +56k$ . $N^{2}-5Nk-75-56k=0$ . $N=\frac{ 5k\pm\sqrt{25k^{2}+224k+300}}{2}$ . Now because N is a positive integer $25k^{2}+224k+300$ must be a perfect square. It is easy to see that $(5k+17)^{2}<25k^{2}+224k+300<(5k+23)^{2}$ .

Now if $25k^{2}+224k+300=(5k+18)^{2}$ . $k=\frac{6}{11}$ which is not an integer. Similarly for $25k^{2}+224k+300=(5k+19)^{2} or (5k+20)^{2} or (5k+21)^{2}$ we don"t get an integer value of k. But for $25k^{2}+224k+300=(5k+22)^{2}$ we get k=46 and therefore $N=\frac{230\pm (5*46 +22)}{2}$ . $N= -11 or N=241$ . but given that N is a positive integer therefore $N=241$ .

I don't know if I was supposed to solve in this way, but I could easily calculate the solution using java:

public static void main(String[] args) {

    int i;

    double a;

    for (i = 0; i <= 1000; i++){

        a = (Math.pow(i, 2) - 75)/(5*i + 56);

        if (a == (int)a){

            System.out.println(i);

        }

    }

}

The output of the program was: 241.

suppose $\frac{N^{2}-75}{5n+56}=y$

$N^{2}-75=5yN+56y$ $N^{2}-5yN-56y-75=0$

$N_{1,2}=\frac{5y\pm \sqrt{25y^{2}+224y+300}}{2}$

suppose $y=2a$ $N_{1,2}=5a\pm \sqrt{25a^{2}+112a+75}$

suppose again $\sqrt{25a^{2}+112a+75}=k$ $25a^{2}+112a+75=k^{2}$ $(k-5a)(k+5a)=112a+75$ from here we can get

$a=23$ and $k=\pm 126$

subtitution to N $N_{1,2}=5a\pm \sqrt{25a^{2}+112a+75}$ $N_{1,2}=5.23\pm 126$ $N_{1}=241$ $N_{2}=-11$ ( Not possible because N is integer )

so the answer is $241$

suppose $\frac{N^{2}-75}{5n+56}=y$

$N^{2}-75=5yN+56y$ $N^{2}-5yN-56y-75=0$

$N_{1,2}=\frac{5y\pm \sqrt{25y^{2}+224y+300}}{2}$

suppose $y=2a$ $N_{1,2}=5a\pm \sqrt{25a^{2}+112a+75}$

suppose again $\sqrt{25a^{2}+112a+75}=k$ $25a^{2}+112a+75=k^{2}$ $(k-5a)(k+5a)=112a+75$ from here we can get

$a=23$ and $k=\pm 126$

subtitution to N $N_{1,2}=5a\pm \sqrt{25a^{2}+112a+75}$ $N_{1,2}=5.23\pm 126$ $N_{1}=241$ $N_{2}=-11$ ( Not possible because N is integer )

so the answer is $241$

(N^2-75)=0.4{(5N+56)(5N-56)+1261} Take, 5N+56=1261 which gives N=241 and it is less than 1000, positive integer. Taking test: (241 241-75)/(5 241+56)=46 So, value of N=241

$\frac{N^{2}-75}{5N+56}$

= $\frac{1}{25}\frac{(5N+56)(5N-56)+1261}{5N+56}$

from this fraction we know that (5N+56)(5N-56) divisible by 5N+56 and 1261 also divisible by 5N+56

1261= $13 \times 97$ = $1 \times 1261$

5N+56 must be equal to positive factor of 1261 because we find N<1000 and N is positive integer.

1. 5N+56=1, N=-11 (negative integer)
2. 5N+56=1261, N=241 (positive integer)
3. 5N+56=13, N= $\frac{-43}{5}$ (not integer)
4. 5N+56=97, N= $\frac{41}{5}$ (not integer)

so possible value of N is 241

Suppose $\frac{N^2-75}{5N+56} = k \in \mathbb Z$ . Rearranging this we get $N^2 - 5kN - 56k = 75$ , and it is easily verified that $(N+56/5)(N-5k-56/5) = 75-(56/5)^2$ . Multiplying both sides by $25$ we have $(5N+56)(5N-25k-56) = -1261 = -13 \times 73$ . So we see that if $N$ is a solution, then if $d = 5N+56$ then $d \mid 13\times 97$ , and then $5 \mid (d-56)$ so $d = 1 \text{ mod 5}$ . Clearly only divisors of $13 \times 97$ with this property are $d = 1$ or $d=1261$ . The first gives us $N = -11$ which is negative, and the second gives us $N = 241$ . Plugging this back into the equation above gives us $k = 46$ . So $\frac{241^2-75}{241*5+56} = 46 \in \mathbb Z$ . Thus the only solution is $\boxed{N = 241}$ .

By partial fraction,

N^2−75/5N+76

if u didn`t solve by partial fraction i will explain.....now m using shortcut

=1/25(5N−56+1261/(5N+56)).

we check that 1261=13⋅97(both primes)

now 5N+56/1261 must be integer so put 5N+56= 1261,

=> N=241 .

Indeed, now further we check put n=241 gives answer 46.....

Therefore 241 is the correct answer

Technique is basically Partial Fraction

Let $k$ be the quotient. Rearranging into quadratic form:

$N^2 - (5k)N + (75 - 56k) = 0$

Using the method of completing the square , i.e. transforming $x^2 + px + q$ into $(x + \frac{p}{2})^2 - \frac{p^2}{4} + q$ , we get:

$(N - \frac{5}{2}k)^2 - \frac{25}{4}k^2 - 75 + 56k = 0$

Multiply by $2^2$ to remove the fractions:

$(2N - 5k)^2 = 25k^2 + 224k + 300$

Now we can complete the square on the right-hand side:

$(2n - 5k)^2 = (5k + 22.4)^2 - (22.4)^2 + 300$

and again multiply through by $5^2$ to remove fractions:

$(10N - 25k)^2 = (25k + 112)^2 - 112^2 + 7500$

Rearranging:

$(25k + 112)^2 - (10N - 25k)^2 = 5044$

Using the identity $a^2 - b^2 = (a+b)(a-b)$ and factorising $5044$ :

$(112 + 10N)(112 - 10N + 50k) = 2*2*13*97$

Note that these two factors differ by $50k - 20N$ , a multiple of 10.

The only factorization of $2*2*13*97$ in which the two factors differ by a multiple of 10 is $2 * 2522$ .

This gives us $112 + 10N = 2522$ , i.e. $N = 241$ .

Final note: This proves that there are no solutions except for potentially $N = 241$ . To complete the proof we do need to actually verify that $N = 241$ solves the problem.

This is easily done and we find $46 = \frac{241^2 - 75}{5*241 + 56}$

Let the given fraction equal some $k\:$ $\in\:$ $\mathbb{Z}$ . Then $N^{2} -75 =\:$ $5Nk + 56k \:,$ or

$N^{2} - 5Nk -75 -56k =\:$ $0$ --- $(i)$ which has integer root(s) for $N$ so it's discriminant is a perfect square i.e.

$25k^{2} + 224k + 300 =\:$ $p^{2}$ --- $(ii)$ for some $p\:$ $\in\:$ $\mathbb{Z^{+}}$ which again has integer root(s) for $k$ so it's discriminant is a perfect square i.e.

$224^{2} - 30000 + 100p^{2} =\:$ $n^{2}$ for some $n\:$ $\in\:$ $\mathbb{Z^{+}}$

$\implies\:$ $(n - 10p)$ $(n + 10p) =\:$ $20176$ --- $(iii)$

Now from -- $(ii)$ we have $k =\:$ $\frac{-224 \pm n}{50}$ --- $(iv)$

Since $k$ is an integer last two digits of $k$ should be one of $24\:,74\:,26\:,76$ . Using this and manually trying out the different possible factorizations of -- $(iii)$ $\big($ There are only three possible factorizations to try out $\big)$ we find $(n - 10p,n + 10p) =\:$ $(4,5044)$ $\implies\:$ $(n,p) =\:$ $(2524,252)$ .

From $(iv),\:$ we get $k =\:$ $46$ $\big($ Rejecting the non-integral root $\big)$ . From -- $(i)$ we obtain $N =\:$ $\boxed{241}$ as our answer.

Using the notation $a|b$ for $a$ divides $b$ , we need to find $N$ where $(5N+56)|N^2-75$ . We use $a|b \Rightarrow a|kb$ and $a|b \Leftrightarrow a|b+ka$ to make this more approachable. by removing all coefficients of $N$ from the right side.

$(5N+56)|N^2-75 \quad \Rightarrow\\ (5N+56)|5(N^2-75)=5N^2-375 \Leftrightarrow\\ (5N+56)|(5N^2-375) - N(5N+56) = -56N-375 \Rightarrow\\ (5N+56)|5(-56N-375) = -280N - 1875 \Leftrightarrow\\ (5N+56)|(-280N-1875) + 56(5N+56) = 56\cdot56 - 1875 = 1261 \\\quad so \\ (5N+56)|1261$

At the stages where the right hand side has been multiplied up by $5$ , the implication only works in one direction. The direction means that every solution of the original is captured in the derived equation, but that not every solution to the derived one is also a solution to the original.

Now we have that $5N+56$ is a factor of $1261 = 13 \times 97$ . The only possible (positive) factors are then $1,13,97,1261$ , and only $1261$ gives a positive integer value of $N$ - $241$ . This has to be checked with the original, and indeed $\frac{241^2-75}{5\cdot241+56} = 46$ .

One note here is that this implies the restriction $N < 1000$ is irrelevant, as this is the only solution possible.

By partial fraction ,N^2−75/5N+76 if u didt solve by partial fraction i will explain.....now m using shortcut =1/25(5N−56+1261/(5N+56)). we check that 1261=13⋅97(both primes) now 5N+56/1261 must be integer so put 5N+56= 1261,=> N=241 .Indeed, now further we check put n=241 gives answer 46.....technique is basically partial fraction

Let $M = \frac{N^2-75}{5N+56}$ be an integer. Then $C = N - 5M$ is also an integer, and we can write $\frac{N^2-75}{5N+56} = \frac{N-C}{5}$ . Cross-multiplying, we get $5N^2-375=5N^2+(56-5C)N-56C$ , so $(56-5C)N=56C-375$ . Since $C$ is an integer, then $56-5C$ is non-zero so we obtain $N = \frac{56C-375}{56-5C}$ . We know that $N$ is positive, so the numerator and denominator must either be both positive or both negative. Since $C$ is an integer, we have $56C-375<0 \Leftrightarrow C \leq 6$ and $56-5C<0 \Leftrightarrow C \geq 12$ . Thus, we cannot have both being negative simultaneously and hence both must be positive. So $7 \leq C \leq 11$ . Also, since $N$ is an integer, we have $(56-5C) | (56C-375)$ and thus $(56 - 5C) | [(56C-375)+11(56-5C)]$ , or $(56 - 5C) | (C + 241)$ . We shall now do casework on $C=7,8,9,10,11$ to see which works. If $C$ is even, then $56-5C$ is even but $C+241$ is odd; so $C$ is odd. If $C=7$ , then $3$ divides $56-5C=21$ but does not divide $C+241=248$ . If $C=9$ , then $56-5C=11$ does not divide $C+241=250$ . The only $C$ left is $11$ , and indeed, $56-5C=1$ divides $11+241=252$ . So $N=\frac{56(11)-375}{56-5(11)}=\boxed{241}$ .

(N^2-75)/(5N+76) is an integer so we can consider (N^2 - 75)/(5N + 56)=a (a is a positive integer)

We have the following equation: N^2 - 75 = a(5N + 56) <=> N^2 - 5aN - 75 - 56a = 0 (1)

This equation has integer roots; therefore Delta must equal b^2 (b is a positive integer)

=> (5a)^2 + 4.56a + 4.75 = b^2 <=> (25a)^2 + 100.56a + 100.75 = (5b)^2 <=> (25a+112)^2 - (5b)^2 = 5044 (2)

By solving (2) we have a=46, b=252 (after eliminating unsatisfied roots)

=> (1) <=> N^2 - 230N - 2651 = 0 <=> N = 241

(N^2-75)/(5N+56)=k, we first multiple by 5 so (5N^2-375)/(5N+56)=5k euclid, 5N^2-375=(5N+56)(N-12)+4N+297 look that 5N+56 not multiplication of 5 so 5N+56=4N+297 then \Box {N=241}

$\frac {n^2-75}{5n+56} \text{ is an integer}\\ \Rightarrow (n^2-75) \vdots (5n+56)\\ \Rightarrow 25(n^2-75) \vdots (5n+56)\\ \Rightarrow (5n+56)^2 - 2 \cdot 56(5n+56) + 1261 \vdots (5n+56)\\ \Rightarrow 1261 \vdots (5n+56)\\\\ \\ \text{ and }1261 = 13 \cdot 97\\ \Rightarrow 5n+56 = 13\\ \text{ or } 5n+56 = 97\\ \text{ or } 5n+56 = 1261\\ \Rightarrow n = 241$

We need $5N + 56 \mid N^2 - 75$ , i.e. there exists $a$ such that $a(5N + 56) = N^2 - 75$ . Write it as quadratic in $N$ to get $$N^2 - 5aN - 56a - 75 = 0$$ This has solution $$N = \frac{\pm \sqrt{25a^2 + 224a + 300} + 5a}{2}$$ So it is necessary to have $$25a^2 + 224a + 300 = x^2$$ for some $x$ . Now note that $$(5a + 23)^2 > 25a^2 + 224a + 300 > (5a + 17)^2$$ i.e. $x$ is equal to one of $5a + 18$ , $5a + 19$ , $\dots$ , $5a + 22$ . It is easily found that only $x = 5a + 22$ yields integral solution $$a = 46$$ Plugging $a = 46$ into $$N = \frac{\pm \sqrt{25a^2 + 224a + 300} + 5a}{2}$$ yields $N = -11$ or $N = 241$ , so $$N = \boxed{241}$$

first. equate it to x, where x is a positive integer..

$N^{2}$ - 75 = 5xN + 56x to equate it to zero, transpose..

$N^{2}$ - 5xN - (75+56x) = 0 using the quadratic formula, [ \frac{[5x +- \sqrt{ \( 25x^{2} + 224x + 300] }{2} ]

we'll now find the values of x that give the discriminant positive.. and from that domain, we'll be able to find that the equation is a perfect square (so that means one solution only) which is 241

5n + 56 | $n^{2}$ - 75

5n + 56 | 5( $n^{2}$ - 75)

5n + 56 | 5 $n^{2}$ + 56n - 56n - 5*75

5n + 56 | n(5n + 56) - 56n - 5 * 75

5n + 56 | -56n - 5 * 75 => 5n + 56 | 56n + 5 * 75

5n + 56 | 5(56n + 5 * 75)

5n + 56 | 5*56n + 56 * 56 - 56 * 56 + 5 * 5 * 75

5n + 56 | 56(5n + 56) + 5 5 75 - 56*56 => 5n + 56 | -1261 => 5n + 56 | 1261

1261 = 13 * 97 => 5n + 56 = 97 or 5n + 56 = 1261

1. 5n + 56 = 97 => 5n = 41 => n is not integer
2. 5n + 56 = 1261=> n = 241

Do long division. You get that $\dfrac{N^2-75}{5N+56}=\dfrac{1}{25}\left(5N-56+\dfrac{1261}{5N+56}\right)$ Now, $1261=13\cdot 97$ , and so the only factor of 1261 that can be expressed as $5N+56$ is 1261, so $N=241$ . This works, so the answer is $\boxed{241}$ .