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Number Theory Level 3

If a ≤ 100 a \le 100 , then how many pairs of positive integers ( y , a ) y,a) satisfy the following?

∑ x = 1 y x 3 = a 2 \large{\sum_{x=1}^{y} {x^{3}} = a^{2}}


The answer is 13.

Department 8 by Department 8 , 27, Israel

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1 solution

Vishrant Goyal
Aug 7, 2015

∑ x = 1 y x 3 = ( y ( y + 1 ) / 2 ) 2 = a 2 \sum_{x=1}^y x^{3} = (y(y+1)/2)^{2} = a^{2}
Since ( y , a ) (y,a) is pair of positive integers and 0 < = y , a < = 100 0<=y,a<=100
a = y ( y + 1 ) / 2........ ( 1 ) a = y(y+1)/2 ........(1)
So, as ( y , a ) (y,a) are positive integers, ( 0 , 0 ) (0,0) is not a solution.
Every value of y from 1..13 satisfies above equation (1) and inequality 0 < = y , a < = 100 0<=y,a<=100
No value of y>13, that satisfies eq(1) satisfies given inequality.
So, ans is 13 \boxed{13}


4 Helpful 0 Interesting 0 Brilliant 0 Confused

This is, by no way, a solution. You need to explain why there are only 13 13 ordered pairs ( y , a ) (y,a) of positive integers ≤ 100 \leq 100 that satisfy the given condition. On a side note, the problem seems overrated to me.

∀ y , a ∈ Z ≤ 100 + , ∑ x = 1 y x 3 = a 2 ⟺ y 2 ( y + 1 ) 2 4 = a 2 ⟺ y ( y + 1 ) 2 = a ⟺ a = T y \begin{aligned}\forall~y,a\in\Bbb{Z^+_{\leq 100}}~,~\sum_{x=1}^y x^3=a^2&\iff \frac{y^2(y+1)^2}{4}=a^2\\&\iff \frac{y(y+1)}2=a\iff a=T_y\end{aligned}

... where T k T_k refers to the k th k^{\textrm{th}} triangular number . The largest triangular number ≤ 100 \leq 100 is T 13 = 91 T_{13}=91 and the smallest triangular number that is positive is T 1 = 1 T_1=1 . Hence, the only solutions are the ordered pairs ( k , T k ) ∀ k ∈ Z ≤ 13 + (k,T_k)~\forall~k\in\Bbb{Z^+_{\leq 13}} giving us 13 13 ordered pairs, hence the answer.

Prasun Biswas - 5 years, 10 months ago

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I'm sorry. I wrote the whole explanation and posted it but it seems my final edit (after all formatting and explanation) didn't come up, I don't know why. So, I'll edit my answer and post it all over again.

Vishrant Goyal - 5 years, 10 months ago

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