Question

Let \[a, b, c, x\] be positive integers such that

\[(x+1)^{4} = 100a + 10b + c\]
and

\[x ^{4} = 100b + 10c + a\]
find

\[(x+1)^{4} - x ^{4}\]

Answer 1

it constitute a number like abc= (x+1)^4
and bca = x^4
so only possible solution I could find is from hit and trial
which gives 4^4 = 256 so a=2 b=5 c=6
and 5^4 = 625 which verifies as bca b=5 c=6 a=2

Answer 2

\[(x+1)^4-x^4 = (2x+1) [(x+1)^2+x^2]\]
\[(x+1)^4-x^4 = 9[11a-10b-c]\]

Answer 3

625-256 = 369

Answer 4

(4+1)^4 = 625 = 100*6 + 10*2 + 5

4^4 = 256 = 100*2 + 10*5 + 6

it matches perfectly hmm

Answer 5

I don't see why there can't be infinitely many solutions to this.

Answer 6

oh sry, it matches pefectly


\(\large \color{black}{\begin{align}
5^4 = \color{red}{6}00+\color{blue}{2}0+\color{brown}{5}\hspace{.33em}\\~\\
4^4 = \color{blue}{2}00+\color{brown}{5}0+\color{red}{6} \hspace{.33em}\\~\\

\end{align}}\)

Answer 7

I'm not saying there are, just that it seems potentially possible since a,b,c are not necessarily less than 10.

Answer 8

if we agree that 100a + 10b+c will be a three digit number then only one solution possible

Answer 9

Sure, there's only one solution that's a 3 digit number. But that's not a very hard problem since there are only a couple numbers when raised the the 4th power that are 3 digit numbers...

Answer 10

In fact, 4 and 5 are the ONLY integers that raise to the fourth power to become 3 digit numbers, so it is trivial to find them.

Answer 11

yeah the problem is solved then... i guess

Answer 12

lets prove there are no solutions when a,b,c > 9