math challenge question sum of squares of three positive odd integers?

656

41^2+43^2+45^2=5555
41*41 on paper =1681
43*43= 1849
45*45=2025
2025+1849+1681=5555
5555=5555

Here is one way to find the answer.  Let x be and odd number.  Then x-2, x, x+2 are three consecutive odd numbers.  The sum of their squares is (x-2)^2 + x^2 + (x+2)^2, which simplifies to 3x^2 + 8.  Also, squares of odd numbers are odd, and the sum of 3 odd numbers is odd.  So we know that 3x^2 + 8 is an odd number.

Now every 4-digit number with identical digits is of the form 1111a, where a is a single digit positive integer, which must be odd to make 1111a an odd number.  So we now know that

3x^2 + 8 = 1111a

where a is one of 1, 3, 5, 7, 9.

Now we see that 3x^2 = 1111a - 8, so the right side 1111a - 8 must be divisible by 3, so adding 9 to it (to make arithmetic easy) must also be divisible by 3.  So now the question is:

1111a + 1 is divisible by 3, for which value of the possible values 1, 3, 5, 7, 9 of a.

a=1 gives 1112, not divisible by 3 (since the sum of its digits is not divisible by 3)
a=3 gives 3334, not divisible by 3 ( same reason)
a=5 gives 5556, is divisible by 3 (since the sum of its digits is 21 which is divisible by 3)
a=7 gives 7778, not divisible by 3
a=9 gives 9999+1 not divisible by 3.

Conclusion:  if this problem has a solution at all then a=5 and 3x^2 + 8 = 5555.
In other words, 5555 is a possible solution and we need to check that it really is a solution.
3x^2 + 8 = 5555 implies that 3x^2 = 5547, so x^2 = 1849, so x^2 = 43^2, so x=43.
So the 3 consecutive odd integers are 41, 43, 45 and sure enough 41^2+43^2+45^2 = 5555.

This question is a little silly, don't you think?

This is not strictly non-calculator, but can be done without a calculaor if you want to spend some time manually.

Let x, x+2, and x+4 be the three consecutive positive integers.
x^2+(x+2)^2+(x+4)^2 = yyyy, where y can be 1 to 9

Simplifying the left-hand side, we have
3x^2+12x+20 = yyyy

You can solve this quadratic with 1111, 2222, 3333,........,9999
5555 gives the following solution.

3x^2+12x+20-5555=0
3x^2+12x-5535=0
This equation is of form ax^2+bx+c
a = 3 b = 12 c = -5535
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[-12 +/-sqrt(12^2-4(3)(-5535)]/(2)(3)
discriminant is b^2-4ac =66564
x=[-12 +?(66564)] / (2)(3)
x=[-12 -?(66564)] / (2)(3)
x=[-12+258] / 6
x=[-12-258] / 6
The roots are 41 and -45
Ignore the negative value:
x=41
The numbers are 41,43, and 45

Please mark as solved.