Problems & Puzzles: Puzzles

 

 

Problems & Puzzles: Puzzles

Puzzle 1245 Even numbers 2m that can be written as a sum of two different numbers x, y whose product xy is a square

On Oct 20, 2025, Emmanuel Vantiahem sent the following interesting puzzle.

Recently, I looked up for even numbers  2m  that can be written as a sum of two different numbers  x, y  whose product  xy  is a square.
Here are the first ones I found :
  10,20,26,30,34,40,50,52,58,60,68,70,74,78,80,82,90,100,102,104,106,110,116,120,122,130,136,140,146,148,150,156,160,164,170,174,178,180,182,190,194,200
As usual I checked this list with sequences in the OEIS.  There was one hit : A072592.

However, that sequence is described as: even numbers that have a prime factor of the form  4w + 1.

So, my question is:

Q. Prove that an even number  2m  is a sum of two different numbers whose product is a square iff m has a prime factor of the form  4w + 1
     or find a counterexample.

Drom 8-14 Nov., 2025, contributions came from Ivan Ianakiev, Emmanuel Vantieghem.

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Ivan wrote:
The sequence A072592 is a subsequence of A337140 (which gives the solution).

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Emmanuel wrote:

Here is my proof :

Assertion : 2m  is the sum of two distinct positive integers whose product is a square <==> 2m  has a prime factor  p  of the form  4w + 1.
Proof of ==> :
Suppose  2m = u + v  with  u*v  square.
Write  u  in the form  s*t^2  where  s  is squarefree and write  v  in the form  x*y^2  with  x  squarefree.
Since  u*v  is a square we have that  s*x*(t*y)^2  is a square.  This is only possible when  s = x.
Thus, 2m = x(t^2 + y^2)  and by a well known theorem of Fermat 
 t^2 + v^2  must have a prime factor of the form  4w + 1.
(see,:  https://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares)
Proof of  <== :
Suppose  2m  has a prime factor  p  of the form  4w + 1.
Then by the same theorem of Fermat, we can write  p  in the form  a^2 + b^2  with  a != b.  
Thus, 2m = k*p = k a^2 + k b^2.  The product of  k a^2  and  k b^2  is  k^2 a^2 b^2; a square.

 

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