Consecutive Odd Integers

An odd integer can always be expressed in the form $2n+1$ for some integer $n$ . Let the two consecutive odd positive integers be denoted by $2n+1$ and $2n+3$ . Then the equation we need to solve is $(2n+3)^3 - (2n+1)^3 = 400 + (2n+1)^2 + (2n+3)^2.$ The expression $(2n+3)^3 - (2n+1)^3$ simplifies to $24 n^2 + 48 n + 26$ , and the expression $400 + (2n+1)^2 + (2n+3)^2$ simplifies to $8n^2 + 16n + 410$ . Subtracting $8n^2 + 16n + 410$ from $24 n^2 + 48 n + 26$ gives us the equation $16n^2 + 32n - 384 = 0$ . We can solve this equation by factoring: $16n^2 + 32n - 384 = 16(n+6)(n-4) = 0$ which gives $n = -6$ or $n = 4$ . When $n = -6$ , the consecutive odd integers are not positive; so the only solution occurs when $n = 4$ . When $n=4$ , the two consecutive odd positive integers are $2n+1 = 9$ and $2n+3 = 11$ , and the sum is $9 + 11 =$ 20.

Since we are dealing with consecutive odd numbers, they can be expressed as $2n-1, 2n+1$ , because $2n$ will always be an even number. We are trying to find $2n+1+2n-1=4n$ (this is the reason why we didn't use $2n+1$ and $2n+3$ , as Arron Kau did: $4n$ is easier to deal with than $4n+4$ ).

From the problem, we know that $(2n+1)^3-(2n-1)^3-(2n+1)^2-(2n-1)^2=400$ .

We shall now take out the common factor, $2n$ . As $(2n+1)^3-(2n-1)^3=(2n+1)^2(2n+1-1)=2n(2n+1)^2$ and $-(2n+1)^2-(2n-1)^2=-(2n-1)^2(2n-1+1)=-2n(2n-1)^2$ , our expression becomes $2n(2n+1)^2-2n(2n-1)^2=400$ . So, we have that $2n((2n+1)^2-(2n-1)^2)=400$ , and because $\color{#D61F06}{(a+x)^2-(a-x)^2=(a+x+a-x)(a+x-a+x)=2a(2x)=4ax}$ , $2n(2n+1+2n-1)(2n+1-2n+1)=2n \times 2 \times 4n=400$ .

Thus, our final answer would be $4n \times 4n=(4n)^2=20^2$ , meaning that our sum ( $4n)$ , is $\sqrt{400}=20$ .

Let the numbers be x and x + 2. (Even solution, if any, can be discarded afterward)

$(x+ 2)^3 - x^3 = 400 + x^2 + (x + 2)^2$ $(x+ 2 - x)[(x+ 2)^2 + x^2 + (x+ 2)x] = 400 + 2x^2 + 4x + 4$ $2(3x^2 + 6x + 4) = 2(200 + x^2 + 2x + 2)$ $x^2 + 2x - 99 = 0$ $(x + 9)(x + 11) = 0$ $x = 9, - 11$

As the numbers are positive, $x = 9, x + 2 = 11$

$9 + 11 = \boxed{20}$

Let x denote the bigger of the two numbers and y the smaller one , then according to the equation $x^{3}-y^{3}=400 + x^{2}+y^{2}》》》 x^{3}-x^{2}= 400 + y^{3}+y^{2}》》》x^{2}(x-1)=400+ y^{2}(y+1)$ Now since x and y are odd consecutive so $x-1 = y+1 =z (let)》》》 x^{2}×z-y^{2}×z=400》》》 z (x^{2}- y^{2})=400》》》 z (x +y)(x-y)= 400 》》》 (y+1)(y+2+y)(y+2-y)=400$ As $x= y+2 》》》 (y+1)^{2}=100$ and y=9 , x=11