- #1
Amcote
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Problem 1
Suppose ab=cd, where a, b, c d [itex]\in[/itex] N. Prove that a[itex]^{2}[/itex]+b[itex]^{2}[/itex]+c[itex]^{2}[/itex]+d[itex]^{2}[/itex] is composite.
Attempt
ab=cd suggests that a=xy, b=zt, c=xz. d=yt. xyzt=xzyt.
So (xy)[itex]^{2}[/itex]+(zt)[itex]^{2}[/itex]+(xz)[itex]^{2}[/itex]+(yt)[itex]^{2}[/itex]=x[itex]^{2}[/itex](y[itex]^{2}[/itex]+z[itex]^{2}[/itex])+t[itex]^{2}[/itex](z[itex]^{2}[/itex]+y[itex]^{2}[/itex])=(x[itex]^{2}[/itex]+t[itex]^{2}[/itex])(z[itex]^{2}[/itex]+y[itex]^{2}[/itex]) Therefore this is composite.
Problem 2
Prove that
GCD(a,b)=1 [itex]\Rightarrow[/itex] GCD(a+b,a-b,ab)=1.
Attempt
My first attempt at this I started with (a+b,a-b,ab)=1 and wrote this as a linear combination (a+b)x+(a-b)y+(ab)z=1 which can be re written as a(x+y+bz)+b(x-y)=1 and I thought I had proved it but realized the arrow suggests a one way proof so now I am stuck at how to start with (a,b)=1... Perhaps if I take ax+by=1 and square both sides but after that I still don't know what to do. Any hints would be appreciated.
Problem 3
Prove that if a,m [itex]\in[/itex] N and a>1, then
GCD([itex]\frac{a^{m}-1}{a-1}[/itex],a-1)=GCD(a-1,m).
Attempt
I'm having a really bad day with these sorts of questions and any attempt just turns into a mess so any little suggestion or hint would be appreciated for this one as well.
Thank you
Suppose ab=cd, where a, b, c d [itex]\in[/itex] N. Prove that a[itex]^{2}[/itex]+b[itex]^{2}[/itex]+c[itex]^{2}[/itex]+d[itex]^{2}[/itex] is composite.
Attempt
ab=cd suggests that a=xy, b=zt, c=xz. d=yt. xyzt=xzyt.
So (xy)[itex]^{2}[/itex]+(zt)[itex]^{2}[/itex]+(xz)[itex]^{2}[/itex]+(yt)[itex]^{2}[/itex]=x[itex]^{2}[/itex](y[itex]^{2}[/itex]+z[itex]^{2}[/itex])+t[itex]^{2}[/itex](z[itex]^{2}[/itex]+y[itex]^{2}[/itex])=(x[itex]^{2}[/itex]+t[itex]^{2}[/itex])(z[itex]^{2}[/itex]+y[itex]^{2}[/itex]) Therefore this is composite.
Problem 2
Prove that
GCD(a,b)=1 [itex]\Rightarrow[/itex] GCD(a+b,a-b,ab)=1.
Attempt
My first attempt at this I started with (a+b,a-b,ab)=1 and wrote this as a linear combination (a+b)x+(a-b)y+(ab)z=1 which can be re written as a(x+y+bz)+b(x-y)=1 and I thought I had proved it but realized the arrow suggests a one way proof so now I am stuck at how to start with (a,b)=1... Perhaps if I take ax+by=1 and square both sides but after that I still don't know what to do. Any hints would be appreciated.
Problem 3
Prove that if a,m [itex]\in[/itex] N and a>1, then
GCD([itex]\frac{a^{m}-1}{a-1}[/itex],a-1)=GCD(a-1,m).
Attempt
I'm having a really bad day with these sorts of questions and any attempt just turns into a mess so any little suggestion or hint would be appreciated for this one as well.
Thank you