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### Four Positive Integers, Any 3 Sum to a Square

```Date: 10/06/2002 at 10:22:32
From: Robert Yalantra
Subject: Four positive integers, any 3 add up to a square of an
integer

Problem: Find four distinct positive integers, a, b, c, and d, such
that each of the four sums a+b+c, a+b+d, a+c+d, and b+c+d is the
square of an integer. Show that infinitely many quadruples (a,b,c,d)
with this property can be created.

I have populated a mySQL database with all possible combinations of
a, b, c, and d with all numbers being between 0 and 30, and do not
find any quadruples that work. Is there an answer to this? Thanks
for any and all help!
```

```
Date: 10/07/2002 at 21:04:07
From: Doctor Paul
Subject: Re: Four positive integers, any 3 add up to a square of an
integer

I've found four and believe that I can produce infinitely many more:

1, 22, 41, 58

9, 34, 57, 78

14, 41, 66, 89

26, 57, 86, 113

While much of my work stemmed from "intelligent" guessing, there was
a bit of mathematics behind my guessing as well.

- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 10/08/2002 at 17:11:24
From: Robert Yalantra
Subject: Four positive integers, any 3 add up to a square of an
integer

I'm very interested in how you did this. I got more quadruples when I
queried the database with larger numbers:

+------+------+------+------+
| a    | b    | c    | d    |
+------+------+------+------+
|    0 |   88 |  168 |  273 |
|    1 |   22 |   41 |   58 |
|    1 |   32 |   88 |  136 |
|    1 |   34 |   65 |  190 |
|    1 |   42 |  153 |  246 |
|    2 |   37 |  130 |  157 |
|    2 |   41 |   78 |  281 |
|    2 |   47 |  207 |  275 |
|    2 |  125 |  162 |  197 |
|    3 |   34 |   63 |  159 |
|    3 |   79 |  114 |  207 |
|    4 |   41 |   76 |  244 |
|    4 |   68 |   97 |  124 |
|    4 |   88 |  164 |  232 |
|    5 |   34 |   61 |  130 |
|    6 |   39 |   99 |  151 |
|    6 |   41 |   74 |  209 |
|    6 |   49 |  201 |  234 |
|    6 |   90 |  129 |  265 |
|    6 |  153 |  241 |  282 |
|    7 |   34 |   59 |  103 |
|    7 |   99 |  183 |  294 |
|    8 |   41 |   72 |  176 |
|    8 |   88 |  160 |  193 |
|    9 |   34 |   57 |   78 |
|    9 |   46 |  114 |  201 |
|    9 |   56 |  224 |  296 |
|    9 |  138 |  177 |  214 |
|   10 |   41 |   70 |  145 |
|   10 |   53 |  133 |  298 |
|   11 |   99 |  179 |  251 |
|   12 |   41 |   68 |  116 |
|   12 |  153 |  196 |  276 |
|   13 |   58 |  218 |  253 |
|   13 |  101 |  142 |  286 |
|   14 |   41 |   66 |   89 |
|   14 |   53 |  158 |  189 |
|   15 |   99 |  175 |  210 |
+------+------+------+------+

I'd love to hear how you did this :-)
Thanks for the response!
```

```
Date: 10/09/2002 at 18:19:35
From: Doctor Paul
Subject: Re: Four positive integers, any 3 add up to a square of an
integer

Here's how I approached the problem:

We want to find integers a, b, c, and d such that

a + b + c = w^2
a + b + d = x^2
a + c + d = y^2
b + c + d = z^2

for some integers w, x, y, and z.

3*(a+b+c+d) = w^2 + x^2 + y^2 + z^2

So w^2 + x^2 + y^2 + z^2 must be a multiple of three.

Now I'm going to use modular arithmetic. If you're not familiar with
modular arithmetic, read one of the entries in our archives.
This might help:

Mod
http://mathforum.org/library/drmath/view/55910.html

Basically, we want to find integers w^2, x^2, y^2, and z^2 such that

w^2 + x^2 + y^2 + z^2 = 0 mod 3.

This can happen in a number of ways:

1. If w^2, x^2, y^2, and z^2 are all congruent to zero mod 3 (ie,
multiples of three), then w^2 + x^2 + y^2 + z^2 = 0 + 0 + 0 + 0 = 0
mod 3.

2. If among the four integers w^2, x^2, y^2, and z^2 we had one
multiple of three and three integers congruent to 1 mod three, then
w^2 + x^2 + y^2 + z^2 = 0 + 1 + 1 + 1 = 0 mod 3.

3. If among the four integers w^2, x^2, y^2, and z^2 we had one
multiple of three and three integers congruent to 2 mod three, then
w^2 + x^2 + y^2 + z^2 = 0 + 2 + 2 + 2 = 0 mod 3.

Now, we can immediately eliminate option three above because the
square of an integer will never be congruent to 2 mod 3 (i.e., 2 is a
quadratic nonresidue mod 3). This is easy to see:

if x is an integer, then there are three possibilities:

x = 0 mod 3
x = 1 mod 3
x = 2 mod 3

Now:

If x = 0 mod 3, then x^2 = 0^2 = 0 mod 3
If x = 1 mod 3, then x^2 = 1^2 = 1 mod 3
If x = 2 mod 3, then x^2 = 2^2 = 4 = 1 mod 3

Thus for all integers x, x^2 will never be congruent to 2 mod 3.

We are still searching for integers w^2, x^2, y^2, and z^2 such that

w^2 + x^2 + y^2 + z^2 = 0 mod 3.

We now know that this can happen in one of these two ways:

1. If w^2, x^2, y^2, and z^2 are all congruent to zero mod 3 (ie,
multiples of three), then w^2 + x^2 + y^2 + z^2 = 0 + 0 + 0 + 0 = 0
mod 3.

2. If among the four integers w^2, x^2, y^2, and z^2 we had one
multiple of three and three integers congruent to 1 mod three, then
w^2 + x^2 + y^2 + z^2 = 0 + 1 + 1 + 1 = 0 mod 3.

Option one may well be valid, but it seemed a bit too "obvious" to
me so I decided to try my luck with option number two.

So I decided to search for squares w^2, x^2, y^2, and z^2 such that
one of them is a multiple of three and the other three are congruent
to 1 mod 3.

I decided to try consecutive squares and see what happened. I figured
this was the easiest case and would quickly be eliminated as a non-
possibility if things weren't working out. But things did in fact work
out. So I guess I got a bit lucky...

If I want to pick four consecutive squares such that only one of them
is a multiple of three, I have to make sure that the first square
isn't a multiple of three. If I did pick the first number to be a
multiple of three, then the fourth number will also be a multiple of
three. To see this, notice that x^2 = 0 mod 3 forces x = 0 mod 3. We
are going to consider the four squares:

x^2, (x+1)^2, (x+2)^2, and (x+3)^2

Notice that if x = 0 mod 3, then (x+3) = 0 mod 3 as well. Then in each
case x^2 and (x+3)^2 will both be congruent to zero mod 3, and this
violates the condition that we only have one multiple of three.

So I have to make sure that the first number is not a multiple of
three.

Before I started all this mathematical analysis, I fiddled with the
numbers a bit and realized that a, b, c, and d couldn't all be
"small."  So I decided to start my guessing with the square 64.
Again I was just guessing. It turns out that I guessed well. :-)

So now I was searching for solutions to:

a + b + c = 64
a + b + d = 81
a + c + d = 100
b + c + d = 121

This is a system of four equations and four unknowns and can be solved
by a variety of methods. Matrices or Gaussian Elimination come to
mind. I used my TI-86 graphing calculator to simplify the work - it
solves linear systems of n equations in n variables.

Solving for a, b, c, and d gives:

a = 1
b = 22
c = 41
d = 58

I decided to see what happened if I chose other sets of consecutive
squares.

I couldn't choose w^2 = 81 because 81 = 0 mod 3.

So my next thought was to try the set

{100, 121, 144, 169}

a = 9
b = 34
c = 57
d = 78

Although I can't prove it, my guess is that choosing x > 7 such that
x = 1 mod 3 or x = 2 mod 3 will always give a solution when you use
the set of squares

{x^2, (x+1)^2, (x+2)^2, (x+3)^2}

The other examples I produced were all obtained in this way. Your list
is much more exhaustive than mine. But it was obtained in an
exhaustive manner.

I hope this sheds some light on the problem. There is still a lot
about the problem that I don't understand. Maybe you can fill in the
blanks...

- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 10/09/2002 at 22:04:56
From: Robert Yalantra
Subject: Thank you (Four positive integers, any 3 add up to a square
of an integer)

Wow, Thank you for all your help! Some of that math was over my head,
but I'm going to print it out and take it to school and see what some
other people think. I really appreciate what you've done; it's leaps
and bounds past my expectations. :-)
```
Associated Topics:
College Number Theory
High School Number Theory

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