Square triangular number: Difference between revisions

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A. V. Sylwester gave a short proof that there are an infinity of square triangular numbers:<ref name=Sylwester>

{{cite journal |last=Pietenpol |first=J. L. |first2=A. V. |last2=Sylwester |first3=Erwin |last3=Just |first4=R. M. |last4=Warten |date=February 1962 |title=Elementary Problems and Solutions: E 1473, Square Triangular Numbers |journal=American Mathematical Monthly |volume=69 |issue=2 |pages=168–169 |ISSN=0002-9890 |jstor=2312558|publisher=Mathematical Association of America | doi = 10.2307/2312558}}

</ref><blockquote> If the {{mvar|n}}th triangular number {{math|{{sfrac|''n''(''n'' + 1)|2}}}} is square, then so is the larger {{math|4''n''(''n'' + 1)}}th triangular number, since:

:<math>\frac{\bigl( 4n(n+1) \bigr) \bigl( 4n(n+1)+1 \bigr)}{2} = 4 \, \frac{n(n+1)}{2} \,\left(2n+1\right)^2.</math>

WeAs know this result has to be a square, because it is athe product of three squares: 4, {{math|{{sfrac|''n''(''n''the +right 1)|2}}}}hand (theside originalis square. triangular number), and {{math|(2''n'' + 1)²}}.</blockquote>The triangular roots {{math|''t_k''}} are alternately simultaneously one less than a square and twice a square if {{mvar|k}} is even, and simultaneously a square and one less than twice a square if {{mvar|k}} is odd. Thus,

We know this result has to be a square, because it is a product of three squares: 4, {{math|{{sfrac|''n''(''n'' + 1)|2}}}} (the original square triangular number), and {{math|(2''n'' + 1)²}}.</blockquote>The triangular roots {{math|''t_k''}} are alternately simultaneously one less than a square and twice a square if {{mvar|k}} is even, and simultaneously a square and one less than twice a square if {{mvar|k}} is odd. Thus,

:49 = 7² = 2 × 5² − 1,

:288 = 17² − 1 = 2 × 12², and