Positive pairs

We can rewrite the given equation as $(a - 1)(a + 1) + 1 + b(a + 1) = 283 \Longrightarrow (a - 1 + b)(a + 1) = 282$ .

Now $a - 1 + b \gt a + 1$ for $b \gt 2$ . If $b = 1$ then the equation becomes $a(a + 1) = 282$ , but as $282 = 2 \times 3 \times 47$ we see that there cannot be two successive integers whose product is $282$ . Also, we cannot have $b = 2$ since the equation would become $(a + 1)^{2} = 282$ , but $282$ is not a perfect square.

Thus it must be the case that $a - 1 + b \gt a + 1$ . Now $282$ can be written as $1 \times 282, 2 \times 141, 3 \times 94$ and $6 \times 47$ . In each of these cases we can then assign the lesser factor to $a + 1$ and the greater factor to $a - 1 + b$ . Since $a \ge 1$ we cannot assign $1$ to $a + 1$ so we can rule out the first of these factorizations. For the others, we find that

• $a + 1 = 2, a - 1 + b = 141 \Longrightarrow a = 1, b = 141$ ,

• $a + 1 = 3, a - 1 + b = 94 \Longrightarrow a = 2, b = 93$ , and

• $a + 1 = 6, a - 1 + b = 47 \Longrightarrow a = 5, b = 43$ .

There are therefore $\boxed{3}$ possible integer solutions $(a,b)$ to the given equation, namely $(1,141), (2, 93)$ and $(5,43)$ .

I solve it with C program

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