How would one find n via congruences (i.e, not by calculating)? $$133^5+110^5+84^5+27^5=n^5$$
I know it's 144, but how do you find it?
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If you are willing to do arithmetic mod $3^6=729$, the left side is $\equiv0$. And then you can conclude $3^2$ divides $n$. And then if you are willing to do arithmetic mod $2^{16}=65536$, you can similarly conclude that $2^4$ divides $n$. Then you would know $144$ divides $n$. – 2'5 9'2 Oct 18 '17 at 02:25
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Working modulo 10, we see $n$ ends in a 4. Working modulo 9, we see that $n$ is a multiple of 3. Working modulo 7, we see $n\equiv4\bmod7$. Also, $n>133$. The first number after 133 to satisfy the congruences is 144; the next is 354, which is clearly too large.
Gerry Myerson
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I'm almost certain this question has been asked and answered here before, but had no luck finding it. – Gerry Myerson Oct 18 '17 at 02:34