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Sums of Odd Numbers Forming a Pythagorean Triple: Find 2b - 2c + a | JEE

JEE Maths question with a full step-by-step solution.

Question
If
(1+3+5+⋯+a)+(1+3+5+⋯+b)=(1+3+5+⋯+c),(1 + 3 + 5 + \cdots + a) + (1 + 3 + 5 + \cdots + b) = (1 + 3 + 5 + \cdots + c),
 such that (i) a+b+c=57a + b + c = 57 and (ii) a<10a < 10, then find the value of (2b−2c+a)(2b - 2c + a).
Solution
Answer: 5
Step 1: Use the standard fact that the sum of the first mm odd numbers is m2m^{2}:
1+3+5+⋯+(2m−1)=m2.1 + 3 + 5 + \cdots + (2m-1) = m^{2} .
 Step 2: Name how many terms each bracket has. Let the three brackets have n1n_1, n2n_2 and n3n_3 terms, so their last terms are
a=2n1−1,b=2n2−1,c=2n3−1.a = 2n_1 - 1, \qquad b = 2n_2 - 1, \qquad c = 2n_3 - 1 .
 Step 3: Rewrite the given equation using Step 1.
n12+n22=n32,n_1^{2} + n_2^{2} = n_3^{2} ,
 so (n1,n2,n3)(n_1, n_2, n_3) is a Pythagorean triple. Step 4: Convert condition (i) into a statement about n1,n2,n3n_1, n_2, n_3.
a+b+c=(2n1−1)+(2n2−1)+(2n3−1)=2(n1+n2+n3)−3=57,a+b+c = \left(2n_1-1\right)+\left(2n_2-1\right)+\left(2n_3-1\right) = 2\left(n_1+n_2+n_3\right) - 3 = 57 ,
  ⟹  n1+n2+n3=30.\implies n_1 + n_2 + n_3 = 30 .
 Step 5: Find the Pythagorean triple with perimeter 3030. Checking the small triples:
(3,4,5)→12,(6,8,10)→24,(5,12,13)→30,(9,12,15)→36.(3,4,5) \to 12, \quad (6,8,10) \to 24, \quad (5,12,13) \to 30 , \quad (9,12,15) \to 36 .
 So {n1,n2}={5,12}\{n_1, n_2\} = \{5, 12\} and n3=13n_3 = 13. Step 6: Use condition (ii) to decide which is which. If n1=12n_1 = 12 then a=23a = 23, breaking a<10a < 10. So
n1=5,n2=12,n3=13.n_1 = 5, \quad n_2 = 12, \quad n_3 = 13 .
 Step 7: Convert back to aa, bb, cc.
a=2(5)−1=9,b=2(12)−1=23,c=2(13)−1=25.a = 2(5)-1 = 9, \qquad b = 2(12)-1 = 23, \qquad c = 2(13)-1 = 25 .
 Check: 9+23+25=579 + 23 + 25 = 57 and 9<109 < 10 . Step 8: Compute the required value.
2b−2c+a=2(23)−2(25)+9=46−50+9=5.2b - 2c + a = 2(23) - 2(25) + 9 = 46 - 50 + 9 = 5 .
 Answer: 55 (i.e. 5.005.00).
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