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Quarting the cube (Posted on 2013-02-28)

Select integer x and triangular number y such that 8y=3x^4-2x^2-1.

Prove that y is divisible by 28 - or find a counter-example.

See The Solution Submitted by broll

Rating: 5.0000 (2 votes)

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SOLUTION +some add-ons

| Comment 3 of 6 |

Counterexample? My dear broll, you must be kidding...

Not only that the "y" s of the (x,y) couples defined by your formula

are divisible by 28,- so are all the "y"s generated by an odd integer x. As long as x mod 7 = 4,6,1, or 3 – no one forces y to be a triang. number.

As a matter of fact 4/7 of the "y"s generated by an odd integer are divisible by 28, 2/7 are divisible by 4 and 1/7 is divisible by 2. -see the table below , showing few "y"s corresponding to odd integers:
(the mod 28 values cycle is 4,20,4, 0,0,0,0, etc )

ymod28<o:p></o:p>

</td> <td width="100" valign="top" style="width:75.0pt;padding:0cm 1.5pt 0cm 1.5pt; height:17.3pt">

y<o:p></o:p>

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x<o:p></o:p>

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</td> </tr> <tr style="mso-yfti-irow:1;height:14.4pt;mso-row-margin-right:74.8pt"> <td width="73" valign="top" style="width:55.1pt;padding:0cm 1.5pt 0cm 1.5pt; height:14.4pt">

0<o:p></o:p>

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0<o:p></o:p>

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1<o:p></o:p>

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</td> </tr> <tr style="mso-yfti-irow:2;height:14.4pt;mso-row-margin-right:74.8pt"> <td width="73" valign="top" style="width:55.1pt;padding:0cm 1.5pt 0cm 1.5pt; height:14.4pt">

0<o:p></o:p>

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28<o:p></o:p>

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3<o:p></o:p>

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4<o:p></o:p>

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228<o:p></o:p>

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5<o:p></o:p>

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</td> </tr> <tr style="mso-yfti-irow:4;height:13.7pt;mso-row-margin-right:74.8pt"> <td width="73" valign="top" style="width:55.1pt;padding:0cm 1.5pt 0cm 1.5pt; height:13.7pt">

20<o:p></o:p>

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888<o:p></o:p>

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7<o:p></o:p>

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4<o:p></o:p>

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2440<o:p></o:p>

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9<o:p></o:p>

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0<o:p></o:p>

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5460<o:p></o:p>

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11<o:p></o:p>

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0<o:p></o:p>

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10668<o:p></o:p>

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13<o:p></o:p>

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0<o:p></o:p>

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18928<o:p></o:p>

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15<o:p></o:p>

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0<o:p></o:p>

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31248<o:p></o:p>

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17<o:p></o:p>

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4<o:p></o:p>

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48780<o:p></o:p>

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19<o:p></o:p>

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20<o:p></o:p>

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72820<o:p></o:p>

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21<o:p></o:p>

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4<o:p></o:p>

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104808<o:p></o:p>

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23<o:p></o:p>

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0<o:p></o:p>

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146328<o:p></o:p>

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25<o:p></o:p>

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0<o:p></o:p>

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199108<o:p></o:p>

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27<o:p></o:p>

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0<o:p></o:p>

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265020<o:p></o:p>

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29<o:p></o:p>

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0<o:p></o:p>

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346080<o:p></o:p>

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31<o:p></o:p>

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4<o:p></o:p>

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444448<o:p></o:p>

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33<o:p></o:p>

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20<o:p></o:p>

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562428<o:p></o:p>

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35<o:p></o:p>

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4<o:p></o:p>

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702468<o:p></o:p>

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37<o:p></o:p>

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0<o:p></o:p>

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867160<o:p></o:p>

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39<o:p></o:p>

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0<o:p></o:p>

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1059240<o:p></o:p>

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41<o:p></o:p>

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0<o:p></o:p>

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1281588<o:p></o:p>

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43<o:p></o:p>

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0<o:p></o:p>

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1537228<o:p></o:p>

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45<o:p></o:p>

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4<o:p></o:p>

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1829328<o:p></o:p>

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47<o:p></o:p>

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</td> </tr> </tbody></table>

As you can notice triangle numbers are a subset of the y column: (0, 28, 5460, 1059240) etc correspond to x column values (1, 3, 11, 41 etc)

For x sequence to continue I have assumed a linear recurrence x(n)= 4*x(n-1)- x(n-2)
(similar to "fun sequence" !!) … and got

x(5)=153 ; x(6) = 459 , compared with Charlie's table, got the corresponding "y" s and assumed a linear recurrence
y(n)= A*y(n-1)+B*y(n-2)+C , then aided by a corresponding applet got the equations constants :
y(n)= 194*y(n-1)1-y(n-2)+28.

Since both y(1) and y(2) i.e. 0 and 28 are divisible by 28, and both 194 and 28

are divisible by 28 - all the triangle numbers in the y sequence will be 0 mod 28.

BTW: all the triangle numbers related to this problem are 4/ 3 of a valid triangle number-

(0 vs 0), (28 vs 21),(5460 vs 4095),

and the x values are such that 3*x^2-2 is a square.

Both the x sequence and y sequence are cited in the OEIS under A001835 & A200999 correspondingly.

.sorry about my format -

<o:p> </o:p>

Posted by Ady TZIDON on 2013-03-01 21:18:12

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