ABCD 5432

First we note that $a$ must be a fourth-power number, and that $c$ must be a square number. So let $a = m^4$ and $c = n^2$ . The equation $c-a = 19 = -(m^2+n)(m^2-n)$ means that, since 19 is a prime number, $m^2+n = 19, m^2-n = -1$ Which means $m=3,n=10$ , which then means $a = 81, c = 100, b = 243, d = 1000$

Because c^3 = d^2 \Rightarrow c^3 is Perfect square \Rightarrow c is a perfect square analogously, a is a perfect square. c - a = 19 \Rightarrow c >= a \Rightarrow \exists k, l \in N such that a = k^2 and c = (k+l)^2 c - a = (k+l)^2 - k^2 = 2kl + l^2 = 19 \Rightarrow l is odd and l^2 <= 19 \Rightarrow l = 1 or l =3 if l = 3 then 6k + 9 = 19 \Rightarrow 6k = 10 (!) if l =1 then 2k + 1 = 19 \Rightarrow k = 9 \Rightarrow a = 81 and c = 100 \Rightarrow b = 243 and d = 1000 so d - b = 1000 - 243 = 757

Suppose a=k^4, b=k^5, c=m^2, d=m^3 for some integer k and m. Since a,b,c,d are postive, then k,m are positive too. Since c-a=19, we have m^2-k^4=19 Factor out, (m+k^2)(m-k^2)=19. Since 19 is prime, its factor are only 1,-1,19,-19. Since m+k^2>0 and m+k^2>m-k^2, we have m+k^2=19 and m-k^2=1. Solving equation, m=10 and k^2=9. k is positive so k=3.

Hence, d-b=m^3-k^5=10^3-3^5=1000-243=757

$a^5=b^4, c^3=d^2$

Thus, there are exist integers $s$ and $t$ such that $a=t^4, b=t^5, c=s^2, d=s^3$ . So, $s^2-t^4=19$ .

$(s-t^2)(s+t^2)=19$

$19$ is prime number and $s+t^2 > s-t^2$

so, $s+t^2=19$ and $s-t^2=1$ .

Then, $s=10, t=3$ and so $d=s^3=1000, b=t^5=243$ and $d-b=\boxed{757}$

Since $a,b,c,d$ are integers, so

$a^5$ = $b^4$

means that $a$ is a perfect fourth power, which also means that it is a perfect square.

Similarly,

$c^3$ = $d^2$

means that $c$ is a perfect square.

So now $c-a$ = 19

This implies that $c$ and $a$ are perfect squares with a difference of 19.

Suppose their square roots are consecutive. Then,

If $c$ = $k^2$ , then $a$ = $(k-1)^2$

So $c-a$ = $2k-1$

Implying that k=10.

So c=10 and a=9.

Substituting these values, we get

b=243

d=1000

So, $d-b$ =757.

Now we show the uniqueness of the solution.

If we substitute values like square of 4, 5, 6, 7, 8, 9 for $c$ then we find that $a$ will not be a perfect square.

So, $c \geq 100$

So we also see that If $c \neq 100$ , then the minimum value of c will become 121.

But the nearest perfect square to 121 is 100,

But $121-100$ =21.

So, $c < 121$ , which proves that c= 10 and a= 9 is a unique solution.

Since $a,b,c$ and $d$ are integers and 4 and 5, as well as 3 and 2, are relatively prime, hence the given equations imply that $a=x^4$ , $b=x^5$ for some positive $x$ , and that $c=y^2$ and $d=y^3$ for some positive integer $y$ . So $c-a=y^2-x^4=(y-x^2)(y+x^2)=19$ . Since 19 is prime, [and the second term is positive] $y-x^2=1, y+x^2=19$ . Solving gives $y=10,x=3$ , and plugging in give $d-b = 757$ .

c and a must be perfect square numbers, so we get c=100, a=81. b=243, d=1000 d-b=757

From $a^5=b^4$ and $c^3=d^2$ we conclude that $a=m^4$ and $c=n^2$ for some positive integer $m,n$ . Then $19=c-a=n^2-m^4=(n+m^2)(n-m^2)$ since $n+m^2\geq 2$ , we get $n+m^2=19$ and $n-m^2=1$ , yield $n=10$ and $m^2=9$ . So, we get $a=81, c=100$ therefore $b=243, d=1000$ and $d-b=757$

Since $a^5 = b^4$ so a must be in the form of $k^4$ for some k

Since $c^3 = d^2$ so c must be in the form of $l^2$ for some l

Looks at k if k=2 ===> a = 32 , c=50 contradiction

if k = 3 ===> a = 81 , c=100 = 10^2 wow yeah !!

so a = 81 c=100 d=1000 b=243 therefore d-b = $757$

To easely solutioning we know if a^5 = b^4 ---> a = p^4 and b = q^5. Then, c^3 = d^2 --> c = r^2 and d = s^3. We know if c - a = 19 and 19 is prime number (19.1) ---> c - a = 19 --> r^2 - p^4 = 19.1 ----> (r + p^2)(r - p^2) = 19.1, so with substitution method we've r = 10 and p =3. So, c = 100 and a = 81. Thus, now we get d = 1000 and b = 243. Finally, d - b = 1000 - 243 = 757. Answer : 757

Lets start by rearranging the equations so as to find positive integer solutions for $a$ and $c$ :

$a^{\frac 54} = b$ and $c^{\frac 32} = d$

According to these rearranged equations, $b$ will have integer values if and only if $a$ is a fourth power number, and $d$ will be an integer if and only if $c$ is a square number. Correspondingly, every fourth power number is in fact a square number, so it is safe to say that $a$ is a square number as well. Now the question states that the difference between $c$ and $a$ is 19. Now if we examine square numbers, we can see that if we have a square number $n^2$ , it can be expressed as a sum of all odd numbers up to $2n - 1$ . The odd number we are aiming to reach through this is 19, and accordingly:

$1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100$

Now we can see that to get 19, we need to subtract $1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81$ from $100$ . As a result, we find $c$ to be $100$ and $a$ to be $81$ . Now to find $d$ and $b$ , we just use the rearranged equations we started with:

$b = 81^{\frac 54} = 243$ and $d = 100^{\frac 32} = 1000$

Therefore, $d - b = 1000 - 243 = \boxed{757}$

The first two equations will be satisfied when a,b,c,d are one's..or if a,b,c and d can all themselves be expressed as powers of a number..in this case the equations are of the form x^4^5=x^5^4 ; y^2^3=y^3^2; where a=x^4; c=y^2...... On trying different values of powers of 4(i.e a) we see that 3^4 satisfies all the conditions...thus d=1000, b=243

We start by looking at $c^{3}$ = $d^{2}$ . Many ordered pairs of positive integers (c,d) satisfy this relationship, for example (4.8), (9,27) and so on. But we can appreciate that such a relationship can only exist if the chosen c is a perfect square.

Now, it is given that $c-a=19$ . So we can say for sure that $c>20$

Also, $a^{5}$ = $b^{4}$ , which again suggests that a must also be a perfect square.

So we look for a $c>20$ such that $c-a=19$ and $a$ is a perfect square. The ordered pair (c,a) is definitely $(100,81)$

Thus, $d=1000$ and $b=243$

Which gives us $d-b=\boxed{757}$

If $a^5=b^4$ then $a=x^4$ and $b=x^5$ . Similarily, $c=y^2$ and $d=y^3$ . So we have $19=c-a=y^2-x^4=(y-x^2)(y+x^2)$ . Since wll terms are positive integers, we have $y-x^2<y+x^2$ and $\{y-x^2,y+x^2\}=\{1,19\}$ . Hence $y-x^2=1$ and $y+x^2=19$ . Solving, we get $y=10, x=3$ . So $d-b=y^3-x^5=1000-243=757$