Perfect Squares!

Question

A question was asked in a mental ability test yesterday.

Write four distinct numbers with these properties:
1. Sum of each pair is a perfect square.
2. Sum of all four is also a perfect squares.

Can you solve this?

Answer 1

Took a long time but worth it.

I got two answers.

Ist set can be:

-88,88,137,488

As:

-88+88 = 0, a perfect square
-88+137 = 49 = 7^2
-88+488 = 400 = 20^2
88+137= 225 =15^2
88+488 = 576 = 24^2
137+488 = 625 =25^2
and finally,
-88+88+137+488=625=25^2

2nd set

-88,344,1177,2792

As:

-88+344 = 256 = 16^2
-88+1177 = 1089 = 33^2
-88+2792 = 2704 = 52^2
344+1177 = 1521 = 39^2
344+2792 = 3136 = 56^2
1177+2792 = 3969 =63^2
and
-88+344+1177+2792=4225 = 65^2

Answer 2

Below is a way to construct many solutions to the problem:

It's easy to see that we can write:
$A = (a^2+b^2-d^2)/2$
$B = (a^2-b^2+d^2)/2$
$C = (d^2+b^2-a^2)/2$
$D = c^2+(d^2-a^2-b^2)/2$
where
$e^2+b^2 = f^2+a^2 = d^2+c^2 = g^2$

So we need square numbers that satisfy this equation, and can then find appropriate values of $A$, $B$, $C$, and $D$. Note that the total number of odd values amongst $a$, $b$, and $d$ must be even (zero or two). So we need to find a square number that can be expressed as a sum of two squares in at least three different ways. We can achieve this by using complex numbers for evaluation.

Consider $125^2$. Now, we can express this in the form $5\times5^2\times5^3$. Noting that $5 = 1^2+2^2$, $5^2=3^2+4^2$, and $5^3=11^2+2^2$, we can find appropriate pairs by evaluating

$±(1+2i)(3±4i)(11±2i)$

which has values $75-100i$, $117+44i$, and $35+120i$ (the fourth case gives 125). So if we let
$a=75$
$b=44$
$c=120$
$d=35$
$e=117$
$f=100$
$g=125$

Then we get
$A = 3168$
$B = 2457$
$C = −1232$
$D = 11232$

Note that we can choose other products, to get different sets of values. For instance, ev3commander's first solution comes from $325^2 = 5^4\times13^2$, which can be written as $325^2 = 5\times65\times325$, with $5=1^2+2^2$, $65=1^2+8^2$, and $325=1^2+18^2$.

ev3commander's second solution comes from $65^2 = 5^2\times13^2 = 5\times13\times65$, with $5=1^2+2^2$, $13=2^2+3^2$, and $65=4^2+7^2$.

wrangler's solution comes from $25^2 = 5\times5\times25$, with $5=1+2^2$ and $25=3^2+4^2$.

Answer 3

It may be

Since it is mental arithmetic and no restrictions.
$3, -3, 4i, -4i$

since:

a) $3 -3 =0$
b) $3 + 4i = (2 + i)(2 + i)$
c) $3 - 4i = (2 - i)(2 - i)$
d) $-3 + 4i = (1 + 2i)(1 + 2i)$
e) $-3 - 4i = (1 - 2i)(1 - 2i)$
f) $4i - 4i = 0$

and also:

$3 - 3 + 4i - 4i =0$

Answer 4

Here we can have...

$a+b=38025=195^2$
$a+c=15625=125^2$
$a+d=8281=91^2$
$c+d=67600=260^2$
$b+d=90000=300^2$ $b+c=97344=312^2$
$a+b+c+d=38025+67600=15625+90000$
$=8281+97344=105625=325^2$

Numbers:

$a= -21847$
$b=59872$
$c=37472$
$d=30128$

seriously no computers, besides posting this. I did all this work on paper. It's possible.

Answer 5

• Before continuing, let me say we have some trivial answers, like $a=b=c=d=0$ or $a=b=0, c=3^2, d=4^2$. But I am looking for an answer with no zero values.

Since we have four integers say, $a$, $b$, $c$, and $d$. We have 6 pairs of them, $$A_1=a+b=A^2\\ A_2=a+c=B^2\\ A_3=a+d=C^2\\ E_3=b+c=D^3\\ E_2=b+d=E^2\\ E_1=c+d=F^2$$

as you may easily see, if $a+b+c+d$ be a square, say $G^2$, the number should be presented as sum of two square in more than 2 different ways.

$$G^2=a+b+c+d=A_1+E_1=A_2+E_2=A_3+E_3$$

Since $5$ and $13$ are the smallest primes of the form $4k+1$, therefor their multiplication, $65$ is the smallest number in which its square could be shown in more than two ways as sum of two squares. so $G=65$, and $G^2=4225$.

Now, lets remember $65^2$ could be written as sum of two square in 4 different ways: $$65^2=4225\\ =256+3969=16^2+63^2\\ =625+3600=25^2+60^2\\ =1089+3136=33^2+56^2\\ =1521+2704=39^2+52^2$$

Since, I am whishing to find smallest answer for the problem, I should find which 3 (out of the 4 possible) pairs of square shall I chose.

Whitout further ado, let me show the answers $$a=-495\\ b=1120\\ c=1584\\ d=2016$$

I found them by solving the following linear equation.

$$\pmatrix{1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 }\times \pmatrix{a \\ b \\ c \\ d}=\pmatrix{A_1 \\ A_2 \\ A_3 \\ 65^2}$$ By selecting $A_1=625$, $A_2=1089$, and $A_3=1521$.

Happy 2016! I bet no one can find a set of 4 nontrivial value each of which have an absolute value smaller than $2016$. ☺

Answer 6

One solution (if these numbers do not have to be positive, but OP said 'no constraints'). Also interpreting 'each pair' as in two separate pairs, i.e. (A, B) and (C,D) but not (A,C) or (B,D):

$-x, x, y, -y$

Then:

$ -x + x = 0 = 0^2$
$ -y + y = 0 = 0^2$
$ -x + x -y + y = 0 = 0^2$
where $x \neq y.$

I am going to see if I can find a more interesting combination, however.

Answer 7

An epic one (kinda small) Also bonus for including 2016?

$a=-495$
$b=2016$ yay
$c= 1120$
$d=1584.$

Squares

$65^2 = 4225=33^2+56^2=25^2+60^2=39^2+52^2$
So, $a+d=1089=33^2$
$b+c=3136=56^2$
$a+c=625=25^2$
$b+d=3600=60^2$
$a+b=39^2=1521$
and $c+d=2704=52^2$.

Answer 8

ANSWER OVERLOAD

-1980, 4480, 8064, 6336. Squares: 2500, 14400, 6084, 10816, 4356, 12544. squares of 50,120,78,104, 66 and 112. Total is 16900 or 130^2

 

-864, 2464, 3465, 2160. Squares are 1600, 5625, 2601,4624, 1296, 5929. Squares of 40;75;51;68;36;77. Total is 7225 or 85^2

 

-3419.5,4644.5; 5365.5, 9044.5. Squares are 1225;14400;1936;13689;5625; 10000. Squares of 35;120;44;117;75; 100. Total is 15625=125^2

Answer 9

Why is everyone using large numbers? A simple solution is:

1, 8, 8, 8

You have sums for pairs:

9 & 16

while for all of the numbers is:

25