Question

The sum of the square of three integers is $324$ . How do you find the integers?

Answer 1

Hint: In this question, we have to find three integers on a given condition. So, we will use the basic mathematical rules to get the solution. We will first find whether the numbers are even or odd numbers. Then, we will find the integers by letting a number such that $\dfrac{324}{3}=108\le {{n}^{2}}\le 324={{18}^{2}}$ , then we solve further and get four different plus-minus sets as the solution for the problem.

Complete step by step solution:
According to the question, we have to find three integers from a given condition.
So, we will use the basic mathematical rules to get the solution.
The condition given to us is that the sum of the square of three integers is $324$ --- (1)
So, let us suppose the number n is an odd number, that is
$odd=n=2k+1$ , where k is some integer
Now, we will take square on both sides in the above equation, we get
${{n}^{2}}={{\left( 2k+1 \right)}^{2}}$
On further solving, we get
${{n}^{2}}=\left( {{\left( 2k \right)}^{2}}+1+2.(2k).(1) \right)$
Therefore, we get
${{n}^{2}}=4{{k}^{2}}+1+4k$
Now, we will take common 4 on the right-hand side of the above equation, we egt
${{n}^{2}}=4({{k}^{2}}+k)+1$
Now, as we know if we add the square of two odd integers, we get an integer of the form $4k+2$ for some integer k.
Also, as per the given condition of the question, the sum of the squares of three numbers is equal to 324, and we know that 324 can be expressed as
$324=4\times 81$
Thus, we see above 324 is expressed as $4k$ instead of $4k+2$ .
Therefore, we get that the three numbers according to the condition are not odd numbers instead they are even numbers.
Also, we will get many solutions for the given problem because ${{n}^{2}}\ge 0$ .
Thus, right now, we will consider the positive integers, and then add a variant involving negative integers at the end.
Let us suppose an integer n such that
$\dfrac{324}{3}=108\le {{n}^{2}}\le 324={{18}^{2}}$
Thus, $12\le n\le 18$
Therefore, possible results for the sum of two integers:
$324-{{18}^{2}}=0$
$324-{{16}^{2}}=68$
$324-{{14}^{2}}=128$
$324-{{12}^{2}}=180$
Let us suppose the largest remaining integer be equal to m, for each values k, thus we get
$\dfrac{k}{2}\le {{m}^{2}}\le k$
Thus we have to get $k-{{m}^{2}}$ to be a perfect square.
Hence, the required result for the solution is
$(0,0,18),\text{ }(2,8,16),\text{ }(8,8,14),\text{ }(6,12,12)$

Therefore, for the condition, the sum of the square of three integers is $324$ , the integers are equal to $(0,0,\pm 18),\text{ }(\pm 2,\pm 8,\pm 16),\text{ }(\pm 8,\pm 8,\pm 14),\text{ }(\pm 6,\pm 12,\pm 12)$

Note: While solving this problem, do mention all the steps properly to avoid confusion and mathematical errors. Always remember to first find whether the integers are positive pr negative. In the end, do not forget to put negative signs as the required solution.