(a) Let p be a prime, and let a be an integer. Show that a is relatively prime to p if and only if there exist integers m and n such that pm + an = 1. (*Hint*) (b) Suppose p is prime, and suppose a is relatively prime to p. Suppose also that p divides ab. By multiplying the equation in part (a) by b, show that p must divide b. (*Hint*) (c) Prove Euclid's Lemma: Let p be a prime number, and let a and b be integers. If p divides ab, then either p divides a or p divides b. (*Hint*)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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The hints for (a) Use Proposition 5.5.16. (b): p must divide the left-hand
side of the multiplied equation (explain why). (c): Consider two cases (I) a
is relatively prime to p; (II) a is not relatively prime to p

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(a) Let p be a prime, and let a be an integer. Show that a is relatively prime
to p if and only if there exist integers m and n such that pm + an = 1.
(*Hint*)
(b) Suppose p is prime, and suppose a is relatively prime to p. Suppose also
that p divides ab. By multiplying the equation in part (a) by b, show
that p must divide b. (*Hint*)
(c) Prove Euclid's Lemma: Let p be a prime number, and let a and b be
integers. If p divides ab, then either p divides a or p divides b. (*Hint*)
Transcribed Image Text:(a) Let p be a prime, and let a be an integer. Show that a is relatively prime to p if and only if there exist integers m and n such that pm + an = 1. (*Hint*) (b) Suppose p is prime, and suppose a is relatively prime to p. Suppose also that p divides ab. By multiplying the equation in part (a) by b, show that p must divide b. (*Hint*) (c) Prove Euclid's Lemma: Let p be a prime number, and let a and b be integers. If p divides ab, then either p divides a or p divides b. (*Hint*)
Proposition 5.5.16. Given the Diophantine equation an + bm = c, where
a, b, c are integers. Then the equation has integer solutions for n and m if
and only if c is a multiple of the gcd of a and b.
PROOF. Since this is an "if and only if" proof, we need to prove it both
ways. We'll do “only if" here, and leave the other way as an exercise.
Since we're doing the "only if" part, we assume that an + bm = c is
solvable. We'll represent the gcd of a and b by the letter d. Since gcd(a, b)
divides both a and b, we may write a
a', b'. By basic algebra, we have an + bm =
this back in the original Diophantine equation, we get:
da' and b = db' for some integers
d(a'n + b'm). If we substitute
d(a'n + b'm) =
It follows that c is a multiple of, d, which is the gcd of a and b.
Transcribed Image Text:Proposition 5.5.16. Given the Diophantine equation an + bm = c, where a, b, c are integers. Then the equation has integer solutions for n and m if and only if c is a multiple of the gcd of a and b. PROOF. Since this is an "if and only if" proof, we need to prove it both ways. We'll do “only if" here, and leave the other way as an exercise. Since we're doing the "only if" part, we assume that an + bm = c is solvable. We'll represent the gcd of a and b by the letter d. Since gcd(a, b) divides both a and b, we may write a a', b'. By basic algebra, we have an + bm = this back in the original Diophantine equation, we get: da' and b = db' for some integers d(a'n + b'm). If we substitute d(a'n + b'm) = It follows that c is a multiple of, d, which is the gcd of a and b.
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