One Step to Squared

Number Theory Level 2

$\begin{array}{c}&&1^2={\color{#D61F06}1}, &0\times2+1={\color{#D61F06}1}\\ &2^2={\color{#D61F06}4}, &1\times3+1={\color{#D61F06}4}\\ &3^2 ={\color{#D61F06}9}, &2 \times 4+1 ={\color{#D61F06}9} \end{array}$

Based on the examples above, is the following true?

For any three consecutive numbers $a, b,$ and $c,$ $b^{2}=ac+1.$

Yes, always No, only sometimes

1 solution

Abygail Spurling
Jun 18, 2018

Substitute $a=(b-1), c=(b+1)$

$b^{2}=(b-1)(b+1)+1$

Multiply $(b-1)(b+1)$

$b^{2}=b^{2}+b-b-1+1$

Simplify

$b^{2}=b^{2}$

This is an interesting relationship which I designed it a few days back :

For any three consecutive numbers $a,b,c$ :

$b^2 = ac + 1$

If you want you can edit your question in this manner. Just a suggestion 😊

Ram Mohith - 2 years, 11 months ago

Thanks for the suggestion! I was struggling a bit with how to word the question concisely without making the answer too obvious.

Abygail Spurling - 2 years, 11 months ago

Also for the examples the following style would be better.

$1^2 = 0 \times 2 + 1 = 1$
$2^2 = 1 \times 3 + 1 = 4$

Examples should reflect theory. As we wrote the condition as $b^2 = a \times c + 1$ the examples should also be in that format. Also highlight the condition in the question as you did in your solution.

Ram Mohith - 2 years, 11 months ago

@Ram Mohith – Thanks for the feedback! This is the first problem I've posted so I appreciate the advice.

Abygail Spurling - 2 years, 11 months ago

Sir, can you please post a solution for this: https://brilliant.org/problems/confusing-question-no-way-out/

Jake Tricole - 1 year, 12 months ago

One Step to Squared

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