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Pirkuliyev Rovsen

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Pirkuliyev Rovsen

#1 h Aug 15, 2014, 5:05 PM • 2 Y

Y by Adventure10, Mango247

Do there exist four distinct integers such that the sum of any two of them is a perfect square?

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mszew

#2 h Aug 15, 2014, 7:23 PM • 2 Y

Y by Adventure10, Mango247

From: http://nrich.maths.org/424

The numbers 2, 34 and 47 are such that the sum of any pair of these numbers is a perfect square.

The integers −208, 224, 352 and 737 also have the property that the sum of any pair of these numbers is a perfect square.

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#3 h Aug 15, 2014, 8:03 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

If you mean positive integers then yes; take $722, 432242, 2814962, 3246482$ . This problem was solved by Euler (see attached).

Attachments:

E797tr.pdf (608kb)

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#4 h Aug 15, 2014, 8:08 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

From http://2000clicks.com/mathhelp/NumberPerfectSquares.aspx.
Positive integers $a,b,c,d$ exist such that the sum of all four of them and the sum of each pair of them are squares.

An example is $386, 2114, 3970, 10430$ . These numbers satisfy

$386+2114=2500=50^2$
$386+3970=4356=66^2$
$386+10430=10816=104^2$
$2114+3970=6084=78^2$
$2114+10430=12544=112^2$
$3970+10430=14400=120^2$
$386+2114+3970+10430=16900=130^2$

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#5 h Aug 16, 2014, 2:42 PM • 2 Y

Y by Adventure10, Mango247

Do there exist four distinct integers such that the sum of any two of them is a perfect square?

This is equivalent to solving the following system of equations:

$\left\{\begin{aligned}& b+a=x^2 \\&b+c=y^2\\&b+f=z^2\\&a+c=e^2\\&a+f=j^2\\&c+f=p^2\end{aligned}\right.$

Let: $F,T,R,D$ - any asked us integers.

For ease of calculation, let's make a replacement.

$q=(8F^2+4FT-T^2)R^2+2(T+2F)RD-D^2$

$k=(8F^2+8FT+2T^2)R^2+2(T+2F)RD$

$s=-T^2R^2+2(T+2F)RD-D^2$

$t=(8F^2+12TF+3T^2)R^2+2(T+2F)DR-D^2$

Then the solutions are of the form:

$x=s^2+k^2-t^2+2(t-k-s)q$

$y=t^2+k^2-s^2+2ks-2tk$

$z=s^2+k^2-t^2$

$e=t^2+k^2+s^2-2kt-2ts$

$j=t^2+s^2-k^2+2ks-2ts$

$p=3s^2+3k^2+3t^2-6kt-6st+8ks+2(t-k-s)q$

$b=\frac{x^2+y^2-e^2}{2}$

$a=\frac{e^2+x^2-y^2}{2}$

$c=\frac{e^2+y^2-x^2}{2}$

$f=\frac{2z^2+e^2-x^2-y^2}{2}$

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