Let
b
1
,
…
,
b
5
b_1, \ldots , b_5
be non-zero integers and
n
n
any integer. Suppose that
b
1
+
⋯
+
b
5
≡
n
(
mod
24
)
b_1+\cdots +b_5 \equiv n \pmod {24}
and
(
b
i
,
b
j
)
=
1
(b_i,b_j)=1
for
1
≤
i
>
j
≤
5
1 \leq i > j \leq 5
. In this paper we prove that (i) if the
b
j
b_j
are not all of the same sign, then the above quadratic equation has prime solutions satisfying
p
j
≪
|
n
|
+
max
{
|
b
j
|
}
25
/
2
+
ε
;
p_j\ll \sqrt {|n|}+ \max \{|b_j|\}^{25/2+\varepsilon };
and (ii) if all the
b
j
b_j
are positive and
n
≫
max
{
|
b
j
|
}
26
+
ε
n \gg \max \{|b_j|\}^{26+\varepsilon }
, then the quadratic equation
b
1
p
1
2
+
⋯
+
b
5
p
5
2
=
n
b_1p_1^2+\cdots +b_5p_5^2=n
is soluble in primes
p
j
.
p_j.
Our previous results are
max
{
|
b
j
|
}
20
+
ε
\max \{|b_j|\}^{20+\varepsilon }
and
max
{
|
b
j
|
}
41
+
ε
\max \{|b_j|\}^{41+\varepsilon }
in place of
max
{
|
b
j
|
}
25
/
2
+
ε
\max \{|b_j|\}^{25/2+\varepsilon }
and
max
{
|
b
j
|
}
26
+
ε
\max \{|b_j|\}^{26+\varepsilon }
above, respectively.