Making the sum of 5th power of integers, a perfect square.

Question

Yesterday this question was posed in a contest. It contains pretty easy questions like asking range of $ab+bc+ca$ when $a^2+b^2+c^2=1$, etc.

But this question is something else. I haven't been able to solve this after $4$ hours.

Question:
Find, $n$ if
$133^5+27^5+84^5+110^5=n^2$.

I checked answer on calculator, it is $248832$.
I tried all sort of things I can do like trying to factor the expression, converting two of the odd nos. into $(m+n)^5+(m-n)^5$ form. But all in vain.

I hope anyone can help me here.

Answer 1

Solution

Let's make a transformation, to find $n \in \mathbb{N_+}$ such that $133^5+27^5+84^5+110^5=m^5.$

Step 1

Estimate the bound of $m$.

Notice that $$133^5<10 \times 100^5,~~~27^5<1 \times 100^5,~~~84^5<1 \times 100^5,~~~110^5<10 \times 100^5.$$Thus $$133^5<m^5=133^5+27^5+84^5+110^5<22 \times 100^5<200^5.$$ Hence, $$133<m<200.$$

Step 2

Find the unit-digit of $m$.

Since $$m^5 \equiv133^5+27^5+84^5+110^5\equiv3^5+7^5+4^5+0^5\equiv3+7+4+0\equiv 4(\textrm{mod }10) ,$$hence the unitdigit of $m$ is $4.$

Step 3

Find $m$.

Since $$m^5 \equiv133^5+27^5+84^5+110^5\equiv1^5+0^5+0^5+2^5\equiv 0(\textrm{mod }3) ,$$hence $3|m^5$. Further, $3|m$. Therefore, $m=144$ or $174.$ However, since$$m^5 \equiv133^5+27^5+84^5+110^5\equiv0^5+6^5+0^5+5^5\equiv 2(\textrm{mod }7) ,$$ and $144^5 \equiv 4^5 \equiv 2 (\textrm{mod }7)$, $174^5 \equiv 6^5 \equiv 6 (\textrm{mod }7)$,$174$ is not the solution. As a result,$$m=144.$$

Answer 2

This is less than a complete solution but too long for a comment.

Suppose $n^2 = a^5 + b^5 + c^5 + d^5$. Then:

$$n^2 = (a+b)m + c^5 + d^5\quad\quad(1)$$

where:

$$m = a^4 -a^3b + a^2b^2 -ab^3 + b^4$$

Applying $(1)$ with $(a,b,c,d) = (133,27,110,84)$:

$$n^2 = 160m + 110^5 + 84^5 = 2^5(5m + 55^5 + 42^5)$$

Hence $2^5 | n^2$ and so $2^3 | n$.

Also, applying $(1)$ with $(a,b,c,d) = (133,110,84,27)$:

$$n^2 = 243m + 84^5 + 27^5 = 3^5(m + 28^5 + 9^5)$$

Hence $3^5 | n^2$ and so $3^3 | n$.

Combining the above results:

$$2^33^3 | n$$