High School Challenge: Find the Positive Integer for $133^5$ + $110^5$ + $84^5$ + $27^5$

[anemone]

Here is this week's POTW:

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Find the positive integer $n$ such that $133^5+110^5+84^5+27^5=n^5$.

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[anemone]

Congratulations to kaliprasad for his correct solution (Cool) , which you can find below:

Spoiler

Let us fix the bound (very rough estimate)

We have $133^5 + 110^5 + 84^5 + 27^5 > 133^5$ so $n>133$
and $133^5 + 110^5 + 84^5 + 27^5 < 4 * 133^5$ so $n < 133 * \sqrt[5]{4} < 133 * \sqrt[4]{4} < 133 * 1.5$

or $n < 200$

so $133 < n < 200$

Now let us work modulo arithmetic

for mod 3 we have

$(133^5+110^5)$ is divisible by 133 + 110 or 243 is it is divisible by 3
$84^5$ and $27^5$ are divisible by 3 so sum is divisible by 3 so n is divisible by 3

for mod 4
$(133^5+27^5)$ is divisible by 133 + 27 or 160 is it is divisible by 4
$84^5$ and $110^5$ are divisible by 4 so sum is divisible by 4 so n is divisible by 4

so n is divisible by 12
or $n \equiv 0 \pmod {12} \cdots(1)$
for mod 5
$(133^5+27^5)$ is divisible by 133 + 27 or 160 is it is divisible by 5
$110^5$ is divisible by 5
$84 \equiv 4 \pmod 5$
raising to power 5 we get
$84^5 \equiv 4 \pmod 5$
so $n^5 \equiv 4 \pmod 5$
or $ n \equiv 4 \pmod 5\cdots(2)$
From (1) by taking multiples of 12 we see that one value 24 satisfies both (1) and (2)
and as 5 and 12 are co-primes we have
$n \equiv 24 \pmod {60}$

we need to find n between 133 and 200 and get n = 144 which satisfies the condition $133 < n < 200$

now we have n = 144 and

$133^5 + 110^5 + 84^5 + 27^5 = 144^5$