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

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Uniteasy Admin · Accepted 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 $$a$$
$$a = \dfrac{n^2-1}{2} =\dfrac{(n-1)(n+1)}{2} $$n
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
$$21^2 = 220+221$$

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