How to find the smallest a, b∈N that solve a single equationa2+a=b2+19b−12

Step 1If a2+ua=b2+vb+wThen 4a2+4ua=4b2+4vb+4w so(2a+u)2−(2b+v)2=u2−v2+4w so(2a+u)2−(2b+v)2=u2−v2+4w oru2−v2+4w=(2a+u+2b+v)(2a+u−2b−v).All solutions come from factoring u2−v2+4w=pq, and solving p=2a+u+2b+v,q=2a+u−2b−vStep 2p+q=4a+2u so a=p+q−2u4 andp−q=4b+2v so b=p−q−2v4.Therefore, u=1,v=19,w=−12u2−v2+4w=1−192−48=−408=−23⋅3⋅17Look at all the factorizations, remembering that p and q can be negative as well as positive and 4 must divide the two expressions.

Expert Answer to "How to find the smallest a, b∈N that..."

High School
Algebra
Algebra I
Quadratic function and equation

NepanitaNesg3a

Answered question

2022-04-21

How to find the smallest a,

b \in N

that solve a single equation

a^{2} + a = b^{2} + 19 b - 12

Answer & Explanation

Friegordigh7r7

Beginner2022-04-22Added 16 answers

Step 1
If

a^{2} + u a = b^{2} + v b + w

Then

4 a^{2} + 4 u a = 4 b^{2} + 4 v b + 4 w

{(2 a + u)}^{2} - {(2 b + v)}^{2} = u^{2} - v^{2} + 4 w

{(2 a + u)}^{2} - {(2 b + v)}^{2} = u^{2} - v^{2} + 4 w

u^{2} - v^{2} + 4 w = (2 a + u + 2 b + v) (2 a + u - 2 b - v)

.
All solutions come from factoring

u^{2} - v^{2} + 4 w = p q

, and solving

p = 2 a + u + 2 b + v, q = 2 a + u - 2 b - v

Step 2

p + q = 4 a + 2 u

a = \frac{p + q - 2 u}{4}

and

p - q = 4 b + 2 v

b = \frac{p - q - 2 v}{4}

.
Therefore,

u = 1, v = 19, w = - 12

u^{2} - v^{2} + 4 w

= 1 - 19^{2} - 48

= - 408

= - 2^{3} \cdot 3 \cdot 17

Look at all the factorizations, remembering that p and q can be negative as well as positive and 4 must divide the two expressions.

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