I'm banging my head against the wall with this task:
Prove that if $x$ is a difference of integer squares, then $3x$ is a difference of integer squares as well.
What strategies could I utilise in order to prove this?
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aachh
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1When is an odd number a difference of squares? What about an even number? – saulspatz Dec 06 '20 at 16:43
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Write $x=u^2-v^2$ with $u,v\in \Bbb Z$, then $3x=3(u^2-v^2)=3(u+v)(u-v)=(3u+3v)(u-v)=((2u+v)+(u+2v))((2u+v)-(u+2v))=(2u+v)^2-(u+2v)^2$
Divide1918
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A little late here but one can generalize the problem as follows:
Claim. If $M,x$ are two integers that can be written as difference of squares of two integers, then $Mx$ is also a difference of squares of two integers. Note that in the original post $M=3=2^2-1^2.$
Proof of claim. Write $M=p^2-q^2,$ and $x=u^2-v^2$, for suitable $u,v,p,q\in\mathbb{Z}$. Then $Mx=(p^2-q^2)(u^2-v^2)=p^2u^2-p^2v^2-q^2u^2+q^2v^2=(pu+qv)^2-(qu+pv)^2.$
ShyamalSayak
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