## A Miracle Equation

I always thought that the famous equation

*10 ^{2} + 11^{2} + 12^{2} = 13^{2} + 14^{2}*

is sort of a miracle, a random fluke. I enjoyed this cute equation, but never really thought about it seriously. Recently, when my son Sergei came home from MOP, he told me that this equation is not a fluke; and I started thinking.

Suppose we want to find five consecutive integers such that the sum of the squares of the first three is equal to the sum of the squares of the last two. Let us denote the middle number by *n*, which gives us the equation:

*(n–2) ^{2} + (n–1)^{2} + n^{2} = (n+1)^{2} + (n+2)^{2}*.

After simplification we get a quadratic equation: *n ^{2} – 12n = 0*, which has two roots, 0 and 12. Plugging

*n = 0*into the equation above gives us

*(–2)*, which doesn’t look like a miracle at all, but rather like a trivial identity. If we replace

^{2}+ (–1)^{2}+ 0^{2}= 1^{2}+ 2^{2}*n*with 12, we get the original miracle equation.

If you looked at how the simplifications were done, you might realize that this would work not only with five integers, but with any odd number of consecutive integers. Suppose we want to find *2k+1* consecutive integers, such that the sum of the squares of the first *k+1* is equal to the sum of the squares of the last *k*. Let us denote the middle number by *n*. Then finding those integers is equivalent to solving the equation: *n ^{2} = 2k(k+1)n*. This provides us with two solutions: the trivial solution 0, and the non-trivial solution

*n = 2k(k+1)*.

So our miracle equation becomes a part of the series. The preceding equation is the well-known Pythagorean triple: *3 ^{2} + 4^{2} = 5^{2}*. The next equation is

*21*. The middle numbers in the series are triangular numbers multiplied by four.

^{2}+ 22^{2}+ 23^{2}+24^{2}= 25^{2}+ 26^{2}+ 27^{2}Actually, do you know that *10 ^{2} + 11^{2} + 12^{2} = 13^{2} + 14^{2} = 365*, the number of days in a year? Perhaps there are miracles or random flukes after all.

## Jonathan:

Also, $latex 20^2 + 22^2 + 24^2 = 26^2 + 28^2 $

20 July 2009, 9:36 pmThere are solutions for longer strings of even consecutive integers.

It looks like there cannot be a solution in odd consecutive integers.

## Tanya Khovanova:

Jonathan, solutions for even numbers can be generated from the solutions for just integers by multiplying them by 2.

20 July 2009, 10:31 pm## Jonathan:

Oops. And I was so pleased with my (n-4)^2 + (n-2)^2 + … work. But yes, it is obvious (with 20/20 hindsight). How did I not notice that I had 20, 22, 24 instead of 10, 11, 12?

21 July 2009, 10:03 am## misha:

Possible generalizations to other degrees (instead of 2) are rather clear.

21 July 2009, 10:57 pm## A nice coincidence «:

[…] nice coincidence By vgtfinal Michael Lugo considers the Pythagorean triple , in which all three numbers are consecutive, and another similar […]

26 April 2010, 5:41 am## Nilotpal Sinha:

I would like to share a similar interesting property with squares. The sum of two integers is a square. The sum of their squares is also a square.

(-3) + 4 = 1² and (-3)² + 4² = 5²

9 + 40 = 7² and 9² + 40² = 41²

This has infinitely many solutions.

2 March 2011, 4:14 am## Alkis Piskas:

Hi,

Check this out about this equation: http://abstrusegoose.com/63

Loved it.

Alkis

3 November 2013, 8:16 am## Habel:

Can anyone answer this???Ram multiplied his year of birth with a big miracle number.His friend Sam also multiplied his year of birth with another miracle number.When they add their results they got an answer equal to 825384558296.What is the year of birth of Sam and Ram.(HINT: eqn used ax+by=ans,where a,b are year of birth(b/w 1920 & 2010) and x,y are big miracle numbers.ans is the sum.)

17 March 2014, 5:12 am