A number theory problem by Wildan Bagus Wicaksono

Number Theory Level 3

Let ${ x }_{ 1 }, { x }_{ 2 }, { x }_{ 3 }, { x }_{ 4 }, ..., { x }_{ 2017 }$ be $2017$ consecutive odd numbers such that ${ x }_{ 1 } + { x }_{ 2 } + { x }_{ 3 } + { x }_{ 4 } + ... +{ x }_{ 2017 }$ is a perfect squared number. Find the smallest value ${ x }_{ 2017 }$ .

The answer is 4033.

1 solution

Wildan Bagus Wicaksono
Jul 3, 2017

$1 = 1^2$

$1 + 3 = 2^2$

$1 + 3 + 5 = 3^2$

$1 + 3 + 5 + 7 = 4^2$

...

$1 + 3 + 5 + ... + (2n - 1) = { \left( \frac { 1+(2n-1) }{ 2 } \right) }^{ 2 }={ n }^{ 2 }$

Thus, the sum of consecutive odd numbers starting from 1 result is always a perfect squared number. Then

${ x }_{ 2017 } = 2 \cdot 2017 - 1 = 4033$ .

So darn I am stupid. I followed your way exactly, but forgot the last step. 2017 isn't the 2017th odd number of course...

Peter van der Linden - 3 years, 11 months ago

Can you prove this by induction?

Zach Abueg - 3 years, 11 months ago

${ \left( \frac { 1+m }{ 2 } \right) }^{ 2 }$ With m is the last digit for $1 + 3 + 5 + ... + m$ and m is odd. It's easier like that.

Wildan Bagus Wicaksono - 3 years, 11 months ago

This is not inductive. You must be able to rigorously prove this for all numbers $n$ ; otherwise, you are just making a conjecture.

Zach Abueg - 3 years, 11 months ago

Yes I know this is not inductive. But this applies if applied. $1+3+5+...+(2(n+1)-1)=1+3+5+...+(2n-1)+(2n+1)\\$

$1+3+5+...+(2(n+1)-1)={ n }^{ 2 }+2n+1\\ 1+3+5+...+(2(n+1)-1)={ (n+1) }^{ 2 }$

Wildan Bagus Wicaksono - 3 years, 11 months ago

A number theory problem by Wildan Bagus Wicaksono

The answer is 4033.

1 solution

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