# 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 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?

• 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. – Ross Millikan Aug 9 at 16:14
• $k^2 + p^2 = 3(a-b)$ 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 $(k-p)(k+p)=k^2 -p^2 = a+b = q^2$ – 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.