Multiples of 4 as sum or difference of 2 squaresIs it true that for any n∈N we can have 4n=x2+y2 or 4n=x2−y2, for x,y∈N∪(0)?I was just working out a proof and this turns out to be true from n=1 to n=20. After that I didn't try, but I would like to see if a counterexample exists for a greater value of n.

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Multiples of 4 as sum or difference of 2 squaresIs it true that for any n∈N we can have 4n=x2+y2 or 4n=x2−y2, for x,y∈N∪(0)?I was just working out a proof and this turns out to be true from n=1 to n=20. After that I didn&#039;t try, but I would like to see if a counterexample exists for a greater value of n.

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Step 1The form x2−y2 factors as (x+y)(x−y). Therefore, if you want to represent an integer N as x2−y2, you can attempt to do so by choosing a factorization N=ab and solving the linear systemx+y=ax−y=bStep 2This system has the unique rational solution x=a+b2,y=a−b2. This gives an integral solution iff a and b have the same parity. Use this to show:A positive integer N is of the form x2−y2 for x,y∈Z (possibly 0) iff N is odd or N is divisible by 4.In particular, 4n is always of the form x2−y2. You want a little more: that x and y are both nonzero. Clearly x cannot be zero, so you need to analyze the case y=0 and show that whenever all possible solutions to n=x2−y2 have y=0, then there are nonzero X and Y such that 4n=X2+Y2.

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