Diophantine Equations (Problem 1)

The problem is just applying this property

$\underbrace{a^{k} + a^{k} + \ldots + a^{k}}_{a \text{ times}}= a^{k+1}$

$a^4 + b^4 + c^4 = d^5$

$a = b = c = d = n$ where $n$ is any natural number

$n^4 + n^4 + n^4 = n^5$

$3n^4 = n^5$

$0 = n^5 - 3n^4$

$0 \equiv n^4(n - 3)$

If $0 \equiv 0$ , substitute $n$ into $3n^4 = n^5$ .

When $n = 1, 1^4(1 - 3) = 1 ( - 2) = -2 \neq 0$

When $n = 2, 2^4 (2 - 3) = 16 (- 1) = -16 \neq 0$

When $n = 3, 3^4 (3 - 3) = 81 (0) = 0 \equiv 0$

Substitute $n = 3$ into $3n^4 = n^5$ :

$3(3^4) = 3^5$

$3(81) = 243$

$243 = 243$

Therefore the second solution where $a, b, c, d$ are integers is $a = b = c = d = 3$