The problem statement from the Ole Miss Problems of the Week page is:

Find three consecutive odd whole numbers such that the sum of their squares is a four-digit number whose digits are all the same.

So we need to find an odd *k* such that

k^{2} + (k+2)^{2} + (k+4)^{2}

is a four-digit number with all four digits the same.

We can program this fairly directly on a programmable calculator, such
as the CFX-9850G or FX-7400G.
We can start *k* at 1, and stop searching when the sum of squares
above exceeds 9999.
The program, available as text here or as a
text file with .CAT file contents,
is:

; Variables: ; A is the k value being tested ; B is the sum of A square, (A+2) square, and (A+4) square ; C is the ones digit of B ; D is the tens digit of B ; E is the hundreds digit of B ; F is the thousands digit of B ; Symbols: ; -> is assignment arrow ; x^2 is square operator ; >= is greater than or equal to relational ; EE is EXP key ; / is division operator ; = is relational operator ; _ is display triangle 1->A ; Start search with 1, the first odd number While 1 ; Loop forever (the exit test is coded inside) Ax^2+(A+2)x^2+(A+4)x^2->B ; Compute the sum of squares If B>=1EE5 ; Test whether the sum of squares is too large Then Break ; If so, exit the loop IfEnd ; [End of test on sum of squares] 10Frac (B/10)->C ; Extract the ones digit of the sum of squares Int (10Frac (B/100))->D; Extract the tens digit Int (10Frac (B/1000))->E; Extract the hundreds digit Int (B/1000)->F ; Extract the thousands digit If C=D And C=E And C=F;Test whether the digits are equal Then A_ ; If so, display the first of the odd numbers IfEnd ; [End of test on equal digits] A+2->A ; Go on to the next odd number WhileEnd ; [End of loop forever] "Done" ; Note program completion

When we run the program, it first reports:

41

and then:

Done

Well, 41^{2} + 43^{2} + 45^{2} = 1681 + 1849 + 2025
= 5555, which indeed is a four digit number with all digits the same.

It is hard, though, to get a feeling of satisfaction from programming a simple search. Perhaps there is another way to do this problem.

We first look at this expression, k^{2} + (k+2)^{2} +
(k+4)^{2}.
Expanding this out, we get

3k^{2} + 12k + 20.

Now how will we express that this sum is a four digit number with all four
digits the same?
Oh, that means the sum is 1111*m* for some integer *m* between
1 and 9 inclusive.
So we have

3k^{2} + 12k + 20 = 1111m, or

3k^{2} + 12k + (20 - 1111m) = 0

Why, this is just a quadratic equation!
Actually, this is nine quadratic equations, depending on the value of
*m*.
We can easily stuff the coefficients into the quadratic formula, and get
results as follows:

m |
k |
---|---|

1 |
-21.1746, +17.1746 |

2 |
-29.1662, +25.1662 |

3 |
-35.2916, +31.2916 |

4 |
-40.4534, +36.4534 |

5 |
-45, 41 |

6 |
-49.1098, +45.1098 |

7 |
-52.8888, +48.8888 |

8 |
-56.4059, +52.4059 |

9 |
-59.7090, +55.7090 |

It is only for *m* equal to 5 that the solutions are integers,
and the solutions are -45 and 41.
The three consecutive odd integers are therefore either 41, 43, and 45,
or -45, -43, and -41.
Clearly these are the same after you square them.
Note that the mathematical solution correctly finds the negative values
that we simply forgot about when doing the program — the problem
did *not* state positive integers, we merely assumed
that.

Oops. Our program cleverly solved the wrong problem. Well, that happens sometimes.

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Copyright © 2001 Brian Hetrick

Page last updated 7 August 2001.