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Prove by contradiction that there are no positive integers p and q such that $$4p^2 - q^2 = 25$$ - Edexcel - A-Level Maths Pure - Question 1 - 2020 - Paper 1

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Prove by contradiction that there are no positive integers p and q such that $$4p^2 - q^2 = 25$$

Worked Solution & Example Answer:Prove by contradiction that there are no positive integers p and q such that $$4p^2 - q^2 = 25$$ - Edexcel - A-Level Maths Pure - Question 1 - 2020 - Paper 1

Step 1

Set up the contradiction and factorise

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Answer

Assume that there exist positive integers p and q such that:

(2p+q)(2pq)=25(2p + q)(2p - q) = 25

Given that 25 is an odd number, both factors must be odd. Therefore, we can express the factors as:

  • Let 2p+q=252p + q = 25 and 2pq=12p - q = 1.

Now, we have two equations to work with.

Step 2

Solve for p and q

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Answer

From the equation 2p+q=252p + q = 25, we can rearrange it to find q:

q=252pq = 25 - 2p

Substituting this expression for q into the second equation:

2p(252p)=12p - (25 - 2p) = 1

Simplifying gives:

2p25+2p=12p - 25 + 2p = 1

4p25=14p - 25 = 1

Thus, we find:

p = 6.5$$ Since p is not a positive integer, this case provides a contradiction.

Step 3

Consider the alternative factorization

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Answer

Now, consider the second case:

  • Let 2p+q=52p + q = 5 and 2pq=52p - q = 5.

From 2p+q=52p + q = 5, we can isolate q:

q=52pq = 5 - 2p

Substituting q into the second equation:

2p(52p)=52p - (5 - 2p) = 5

This simplifies to:

2p5+2p=52p - 5 + 2p = 5

4p = 10 \ p = 2.5$$ Again, since p is not a positive integer, we reach another contradiction.

Step 4

Conclusion

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Answer

Since both cases lead us to a contradiction where p is not a positive integer, we conclude that:

There are no positive integers p and q such that

4p2q2=254p^2 - q^2 = 25

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