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Five consecutive natural numbers with sum n^3 and middle three sum m^2 | JEE Main

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Question
If x1,x2,x3,x4,x5x_1,x_2,x_3,x_4,x_5 are five consecutive natural numbers with
x1+x2+x3+x4+x5=n3andx2+x3+x4=m2,x_1+x_2+x_3+x_4+x_5 = n^3 \qquad\text{and}\qquad x_2+x_3+x_4 = m^2 ,
 where mm and nn are natural numbers, then the least possible value of n3+m22\dfrac{n^3+m^2}2
Solution
Answer: 2700
Step 1: Let x3=kx_3 = k, so the numbers are k−2, k−1, k, k+1, k+2k-2,\ k-1,\ k,\ k+1,\ k+2 with k≥3k \ge 3.
x1+x2+x3+x4+x5=5k=n3...(1)x_1 + x_2 + x_3 + x_4 + x_5 = 5k = n^3 \quad ...(1)
x2+x3+x4=3k=m2...(2)x_2 + x_3 + x_4 = 3k = m^2 \quad ...(2)
∴ n3+m22=5k+3k2=4k\therefore\ \frac{n^3 + m^2}{2} = \frac{5k + 3k}{2} = 4k
 So the least value is obtained at the least kk. Step 2: From (1), 5∣n35 \mid n^3. 55 is prime, so 5∣n5 \mid n. Let n=5an = 5a, a∈Na \in N.
5k=(5a)3=125a3  ⟹  k=25a35k = (5a)^3 = 125a^3 \implies k = 25a^3
 Step 3: Putting in (2),
m2=3k=75a3=52×3a3m^2 = 3k = 75a^3 = 5^2 \times 3a^3
 m2m^2 and 525^2 are perfect squares, so 3a33a^3 is a perfect square. Step 4: Taking a=1,2,3a = 1, 2, 3 in order,
a=1: 3a3=3,a=2: 3a3=24a = 1:\ 3a^3 = 3, \qquad a = 2:\ 3a^3 = 24
 neither is a perfect square, and
a=3: 3a3=81=92a = 3:\ 3a^3 = 81 = 9^2
 ∴\therefore the least value of aa is 33. Step 5: a=3  ⟹  n=15a = 3 \implies n = 15, k=25×27=675k = 25 \times 27 = 675, m2=3×675=2025  ⟹  m=45m^2 = 3 \times 675 = 2025 \implies m = 45. The numbers are 673,674,675,676,677673, 674, 675, 676, 677, all natural. Check: 5k=3375=1535k = 3375 = 15^3 and 674+675+676=2025=452674 + 675 + 676 = 2025 = 45^2. Step 6:
n3+m22=4k=4×675=2700\frac{n^3 + m^2}{2} = 4k = 4 \times 675 = 2700
 Answer: 27002700.
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