Question

Express the following squares as the sum of two consecutive integers.

11^2

13^2

17^2

21^2

Collected in the board: Number Theory

Steven Zheng posted 8 hours ago

Answer

If a perfect square is an odd number, it can be expressed as the sum of two consecutive integers.

Let n^2 represent the perfect square, a as the smaller one of the two consecutive integers. Then, a+1 is the bigger one. And we have the following equation.

n^2 = a+(a+1)

Solve for a

a = \dfrac{n^2-1}{2} =\dfrac{(n-1)(n+1)}{2}
(1)

then

a+1 = \dfrac{n^2-1}{2} +1 = \dfrac{n^2+1}{2}

Therefore

n^2 can be represented by the two consecutive integers below.

n^2 = \dfrac{n^2-1}{2} + \dfrac{n^2+1}{2}

11^2

n = 11

a = \dfrac{11^2-1}{2} =\dfrac{(11-1)(11+1)}{2} =60

Then

11^2 = 60+61

13^2

n=13

a = \dfrac{13^2-1}{2} =\dfrac{(13-1)(13+1)}{2} =84

Then

13^2 = 84+85

17^2

n=17

a = \dfrac{17^2-1}{2} =\dfrac{(17-1)(17+1)}{2} =144

Then

17^2 = 144+145

21^2

n = 21

a = \dfrac{21^2-1}{2} =\dfrac{(21-1)(21+1)}{2} =220

Then

21^2 = 220+221

Steven Zheng posted 7 hours ago

Scroll to Top