Let a and b be natural numbers such that $2a - b, a - 2b$ and $a + b$ are all distinct squares. What is the smallest possible value of $b$?

Let, $2a-b=k^2, a-2b=p^2, a+b=q^2$.

$k^2=p^2+q^2$ after adding any of the two equations.

How to proceed further?

  • 3
    $\begingroup$ Differences of squares are more interesting because they can be factored. Alternately, you could just try some of the small Pythagorean triangles you know and see if they work. $\endgroup$ – Ross Millikan Aug 9 at 16:14
  • $\begingroup$ $k^2 + p^2 = 3(a-b)$ which might be interesting. Or might not. $(k^2+p^2)q^2 = 3(a^2 - b^2)$. Interesting looking but might not pan out to much. But definitely follow Ross Millikan's advice and note $(k-p)(k+p)=k^2 -p^2 = a+b = q^2$ $\endgroup$ – fleablood Aug 9 at 16:32

If you subtract the last two you get $q^2-p^2=3b$. If you add the first and last you get $k^2+q^2=3a$. No primitive Pythagorean triangle has legs that differ by a multiple of $3$, so we need a triangle that has a common factor of $3$. The smallest such is $9-12-15$ and we find $$3b=144-81=63\\b=21\\3a=225+144=369\\a=123$$ This is the smallest $b$ because the difference of the two legs must be at least $3$. If the shorter leg is $c$ we have $(c+3)^2-c^2=6c+9$ and $b$ will grow with the shorter leg.


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