Question

$(a + 22)$
is a square number.
$(a - 23)$
is another square number. Find the value of a.

​

Answer 1

Let there exists two integers $p$ and $q$ such that,

$\longrightarrow a+22=p^2$

$\longrightarrow a=p^2-22\quad\quad\dots(1)$

And,

$\longrightarrow a-23=q^2$

$\longrightarrow a=q^2+23\quad\quad\dots(2)$

Subtracting (1) from (2),

$\longrightarrow a-a=(p^2-22)-(q^2+23)$

$\longrightarrow 0=p^2-22-q^2-23$

$\longrightarrow 0=p^2-q^2-45$

$\longrightarrow p^2-q^2=45$

$\longrightarrow (p+q)(p-q)=45$

Now we assume some possible values for $p+q$ and $p-q$ as cases.

Case 1:- Let $p+q=45$ and $p-q=1.$

Then we get,

$\longrightarrow p=\dfrac{45+1}{2}=23$

$\longrightarrow q=\dfrac{45-1}{2}=22$

Hence, from (2),

$\longrightarrow a=22^2+23$

$\longrightarrow\underline{\underline{a=507}}$

Case 2:- Let $p+q=15$ and $p-q=3.$

Then we get,

$\longrightarrow p=\dfrac{15+3}{2}=9$

$\longrightarrow q=\dfrac{15-3}{2}=6$

Hence, from (2),

$\longrightarrow a=6^2+23$

$\longrightarrow\underline{\underline{a=59}}$

Case 3:- Let $p+q=9$ and $p-q=5.$

Then we get,

$\longrightarrow p=\dfrac{9+5}{2}=7$

$\longrightarrow q=\dfrac{9-5}{2}=2$

Hence, from (2),

$\longrightarrow a=2^2+23$

$\longrightarrow\underline{\underline{a=27}}$

Hence the possible values of $a$ are taken in a set as,

$\longrightarrow\underline{\underline{a\in\{27,\ 59,\ 507\}}}$

Let's verify for each.

For $a=27,$

$\longrightarrow 27-23=4=2^2$

$\longrightarrow 27+22=49=7^2$

For $a=59,$

$\longrightarrow 59-23=36=6^2$

$\longrightarrow 59+22=81=9^2$

For $a=507,$

$\longrightarrow 507-23=484=22^2$

$\longrightarrow 507+22=529=23^2$