HCF of polynomials
$begingroup$
If hcf(P,Q)=hcf(P,R)=1 then hcf(P,QR)=1 ,how would I try a prove this, I tried using Bezout's Lemma to get APR-R=-B(QR), but how would I proceed?
elementary-number-theory
asked Jan 2 at 11:43
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
$endgroup$
add a comment |
$begingroup$
If hcf(P,Q)=hcf(P,R)=1 then hcf(P,QR)=1 ,how would I try a prove this, I tried using Bezout's Lemma to get APR-R=-B(QR), but how would I proceed?
elementary-number-theory
asked Jan 2 at 11:43
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
$endgroup$
add a comment |
$begingroup$
If hcf(P,Q)=hcf(P,R)=1 then hcf(P,QR)=1 ,how would I try a prove this, I tried using Bezout's Lemma to get APR-R=-B(QR), but how would I proceed?
elementary-number-theory
asked Jan 2 at 11:43
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
$endgroup$
If hcf(P,Q)=hcf(P,R)=1 then hcf(P,QR)=1 ,how would I try a prove this, I tried using Bezout's Lemma to get APR-R=-B(QR), but how would I proceed?
elementary-number-theory
elementary-number-theory
asked Jan 2 at 11:43
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
asked Jan 2 at 11:43
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
asked Jan 2 at 11:43
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
asked Jan 2 at 11:43
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
asked Jan 2 at 11:43
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
386
386
add a comment |
add a comment |
1 Answer
1
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$begingroup$
One possibility, using Bézout, is to find $s, t, u, v$ such that
$$
s P + t Q = 1 = u P + v R,
$$
and then compute
$$
1 = s P + t Q = s P + t Q cdot 1 = s P + t Q (u P + v R) = dots
$$
answered Jan 2 at 11:51
Andreas CarantiAndreas Caranti
56.6k34395
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add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One possibility, using Bézout, is to find $s, t, u, v$ such that
$$
s P + t Q = 1 = u P + v R,
$$
and then compute
$$
1 = s P + t Q = s P + t Q cdot 1 = s P + t Q (u P + v R) = dots
$$
answered Jan 2 at 11:51
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
add a comment |
$begingroup$
One possibility, using Bézout, is to find $s, t, u, v$ such that
$$
s P + t Q = 1 = u P + v R,
$$
and then compute
$$
1 = s P + t Q = s P + t Q cdot 1 = s P + t Q (u P + v R) = dots
$$
answered Jan 2 at 11:51
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
add a comment |
$begingroup$
One possibility, using Bézout, is to find $s, t, u, v$ such that
$$
s P + t Q = 1 = u P + v R,
$$
and then compute
$$
1 = s P + t Q = s P + t Q cdot 1 = s P + t Q (u P + v R) = dots
$$
answered Jan 2 at 11:51
Andreas CarantiAndreas Caranti
56.6k34395
$endgroup$
One possibility, using Bézout, is to find $s, t, u, v$ such that
$$
s P + t Q = 1 = u P + v R,
$$
and then compute
$$
1 = s P + t Q = s P + t Q cdot 1 = s P + t Q (u P + v R) = dots
$$
answered Jan 2 at 11:51
Andreas CarantiAndreas Caranti
56.6k34395
answered Jan 2 at 11:51
Andreas CarantiAndreas Caranti
56.6k34395
answered Jan 2 at 11:51
Andreas CarantiAndreas Caranti
56.6k34395
answered Jan 2 at 11:51
Andreas CarantiAndreas Caranti
56.6k34395
56.6k34395
add a comment |
add a comment |
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