Question

Доказать, что ни при каком целом n число n^2+5n+16 не делится на 169.

Answer 1

n^2+5n+16=(n-4)^2+13n

2 слагаемое делится на 13, значит нужно что бы 1 делилось на 13, а это значит n-4 тоже должно делиться на 13.значит все 1 слагаемое делится на 169. и значит 2 тоже должно делится на 169, а это значит, что n должно делиться на 13. А это противоречит 1 слагаемому. т.к. n-4 и n не могут одновременно делиться на 13, значит все выражение не делиться на 169

Answer 2

Proof that n^2 + 5n + 16 is not divisible by 169

To prove that the expression n^2 + 5n + 16 is not divisible by 169 for any integer n, we can use a proof by contradiction.

Assume that there exists an integer n for which n^2 + 5n + 16 is divisible by 169. This means that there exists another integer k such that:

n^2 + 5n + 16 = 169k

Now, let's rearrange the equation:

n^2 + 5n + (16 - 169k) = 0

This is a quadratic equation in terms of n. We can solve this equation using the quadratic formula:

n = (-5 ± √(5^2 - 4(16 - 169k)))/(2)

Simplifying further:

n = (-5 ± √(25 + 4(169k - 16)))/(2)

n = (-5 ± √(25 + 676k - 64))/(2)

n = (-5 ± √(676k - 39))/(2)

Since n is an integer, the discriminant (676k - 39) must be a perfect square. Let's denote the discriminant as m^2:

676k - 39 = m^2

Rearranging the equation:

676k = m^2 + 39

Now, let's consider the possible values of m^2 modulo 13. We know that 676 is divisible by 13, so the left side of the equation is congruent to 0 modulo 13. Therefore, m^2 + 39 must also be congruent to 0 modulo 13.

m^2 + 39 ≡ 0 (mod 13)

Subtracting 39 from both sides:

m^2 ≡ -39 (mod 13)

Simplifying further:

m^2 ≡ -1 (mod 13)

Now, let's consider the quadratic residues modulo 13. The quadratic residues modulo 13 are 0, 1, 3, 4, 9, and 10. None of these residues are congruent to -1 modulo 13. Therefore, there are no integer solutions for m^2 ≡ -1 (mod 13).

Since there are no integer solutions for m^2 ≡ -1 (mod 13), there are no integer solutions for 676k = m^2 + 39. This means that there are no integer solutions for n^2 + 5n + 16 = 169k.

Therefore, we have proved that n^2 + 5n + 16 is not divisible by 169 for any integer n.