On the Diophantine Equation nx+ 10y=z2
SUTON TADEE
Department of Mathematics,
Faculty of Science and Technology,
Thepsatri Rajabhat University, Lopburi 15000,
THAILAND
Abstract: In this paper, we show that (n, x, y, z) = (2,3,0,3) is the unique non-negative integer solution of
the Diophantine equation nx+ 10y=z2, where nis a positive integer with n≡2 (mod 30) and x, y, z are
non-negative integers. If n= 5, then the Diophantine equation has exactly one non-negative integer solution
(x, y, z) = (3,2,15). We also give some conditions for non-existence of solutions of the Diophantine equation.
Key-Words: Diophantine equation, Mih˘ailescu’s Theorem, congruence, non-negative integer solution
Received: October 29, 2022. Revised: December 24, 2022. Accepted: January 22, 2023. Published: February 23, 2023.
1 Introduction
In 2014, Sroysang, [1], proved that the Diophantine
equation 4x+ 10y=z2has no non-negative integer
solution. After that, in 2019, Burshtein, [2], showed
that the Diophantine equation 7x+ 10y=z2has
no positive integer solution. In 2020, Orosram and
Comemuang, [3], found that the Diophantine equa-
tion 8x+ny=z2, where nis a positive integer
with n≡10 (mod 15), has the unique non-negative
integer solution (x, y, z) = (1,0,3). In 2021, N.
Viriyapong and C. Viriyapong, [4], proved that the
Diophantine equation nx+ 13y=z2, where nis a
positive integer with n≡2 (mod 39) and n+ 1 is
not a square number, has the unique non-negative in-
teger solution (n, x, y, z) = (2,3,0,3). Tangjai and
Chubthaisong, [5], studied the Diophantine equation
3x+py=z2, where pis prime and p≡2 (mod 3)
and found that if y= 0, then (p, x, y, z) = (p, 1,0,2)
is the only one non-negative integer solution and if 4∤
y, then the equation has the unique non-negative inte-
ger solution (p, x, y, z) = (2,0,3,3). In 2022, Wan-
naphan and Tadee, [6], found all non-negative integer
solutions of the Diophantine equation n2x+ 2y=z2,
where nis an odd positive integer. In the same year,
N. Viriyapong and C. Viriyapong, [7], proved that
the Diophantine equation nx+ 19y=z2, where n
is a positive integer with n≡2 (mod 57) has the
unique non-negative integer solution (n, x, y, z) =
(2,3,0,3). Borah and Dutta, [8], studied the Dio-
phantine equation nx+24y=z2, where n is a positive
integer with n≡5,7 (mod 8).
Inspired by the work mentioned earlier, we study
the Diophantine equation nx+ 10y=z2, where n
is a positive integer. We can easily notice that if
n≡1( mod 3), then the Diophantine equation has no
non-negative integer solution. Since n≡1 (mod 3),
we have z2=nx+ 10y≡2 (mod 3), a contradic-
tion since z2≡0,1 (mod 3). Cases that have not yet
been considered, are n≡0,2 (mod 3). In this re-
search, we will consider the case n≡2 (mod 3) and
n≡2 (mod 10). That is n≡2 (mod 30). More-
over, we study in case n= 5.
2 Preliminaries
In the beginning this section, we present some helpful
Theorems.
Theorem 1. If zis an integer, then z2≡0,1,4,
5,6,9 (mod 10).
Proof. Let zbe an integer. Then there exists a
non-negative integer rsuch that z≡r(mod 10),
where 0≤r≤9.
Case 1: r= 0. Then z2≡0 (mod 10).
Case 2: r= 1. Then z2≡1 (mod 10).
Case 3: r= 2. Then z2≡4 (mod 10).
Case 4: r= 3. Then z2≡9 (mod 10).
Case 5: r= 4. Then z2≡16 ≡6 (mod 10).
Case 6: r= 5. Then z2≡25 ≡5 (mod 10).
Case 7: r= 6. Then z2≡36 ≡6 (mod 10).
Case 8: r= 7. Then z2≡49 ≡9 (mod 10).
Case 9: r= 8. Then z2≡64 ≡4 (mod 10).
Case 10: r= 9. Then z2≡81 ≡1 (mod 10).
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Theorem 2. [9], (x, y, z)∈ {(3,0,3),(2,1,3)}are
exactly two non-negative integer solutions of the
Diophantine equation 2x+ 5y=z2.
Theorem 3. (Mih˘ailescu’s Theorem), [10], The
Diophantine equation ax−by= 1 has the unique
solution (a, b, x, y) = (3,2,2,3), where a, b, x, y are
integers and min{a, b, x, y}>1.
Next, we prove two useful Lemmas by using
Mih˘ailescu’s Theorem.
Lemma 4. The Diophantine equation
1 + 10y=z2(1)
has no non-negative integer solution.
Proof. Assume that (y, z)is a non-negative integer
solution of (1). Then z2−10y= 1. It is easy to check
that y > 1and z > 1. Thus min {z, 10,2, y}>1. By
Theorem 3, this is impossible.
Lemma 5. Let nbe a positive integer with n≡
2 (mod 10). Then the Diophantine equation
nx+ 1 = z2(2)
has a unique non-negative integer solution. The solu-
tion is (n, x, z) = (2,3,3).
Proof. Let (x, z)be a non-negative integer solution of
(2). If n= 1 or x= 0, then z2= 2, a contradiction.
Thus n > 1and x≥1. If x= 1, then n+ 1 = z2.
Since n≡2 (mod 10), we have z2≡3 (mod 10).
This is impossible by Theorem 1. Then x > 1. Next
we consider z. If z= 0 or z= 1, then nx=−1
or nx= 0, respectively, a contradiction. Thus z > 1
and so min{z, n, 2, x}>1. By Theorem 3 and (2),
we have (n, x, z) = (2,3,3).
3 Main Results
In this section, we give our results.
Theorem 6. The Diophantine equation
5x+ 10y=z2(3)
has a unique non-negative integer solution. The solu-
tion is (x, y, z) = (3,2,15).
Proof. Let (x, y, z)be a non-negative integer solution
of (3). Suppose that x≥y. From (3), we have
5y(5x−y+ 2y) = z2.(4)
Then yis even and there exists a positive integer z1
such that
5x−y+ 2y=z2
1.(5)
By Theorem 2, we have y= 2 and x−y= 1.
Then x= 3, and so z2= 53+ 102= 225. Hence
(x, y, z) = (3,2,15) is a non-negative integer solu-
tion of (3). Now, we consider x<y. From (3), it
follows that
5x(1 + 2y·5y−x) = z2.(6)
Thus, xis even and there exists a positive integer z2
such that 1+2y·5y−x=z2
2. It implies that
(z2−1)(z2+ 1) = 2y·5y−x.(7)
Then there exists two non-negative integers uand v
such that
z2−1 = 2u·5v(8)
and
z2+ 1 = 2y−u·5y−x−v.(9)
From (8) and (9), we get
2 = 2y−u·5y−x−v−2u·5v.(10)
Now, we consider three following cases:
Case 1: y−x−v= 0. From (10), we obtain that
2 = 2y−u−2u·5v.(11)
Subcase 1.1: y−u≥u. From (11), we ob-
tain that 2 = 2u(2y−2u−5v). Then u= 1 and
1=2y−2u−5v. It is easy to check that y−2u > 1
and v > 1. This is impossible by Theorem 3.
Subcase 1.2: y−u<u. From (11), we get
2 = 2y−u(1−22u−y·5v). Then y−u= 1 and
1=1−22u−y·5v. Thus 22u−y·5v= 0, a contra-
diction.
Case 2: v= 0. From (10), we obtain that
2 = 2y−u·5y−x−2u.(12)
Subcase 2.1: y−u≥u. From (12), we get
2=2u(2y−2u·5y−x−1). Then u= 1 and
1 = 2y−2u·5y−x−1. Thus 2y−2u·5y−x= 2, and so
y−x= 0. This is impossible since x < y.
Subcase 2.2: y−u < u. From (12), we get
2 = 2y−u(5y−x−22u−y). Then y−u= 1 and
5y−x−22u−y= 1. It is easy to check that y−x > 1
and 2u−y > 1. This is impossible by Theorem 3.
Case 3: y−x−v > 0and v > 0. From (10), we get
5|2, a contradiction.
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Theorem 7. Let nbe a positive integer with n≡
2 (mod 30). Then the Diophantine equation
nx+ 10y=z2(13)
has a unique non-negative integer solution. The solu-
tion is (n, x, y, z) = (2,3,0,3).
Proof. Let x, y and zbe non-negative integers such
that the equation (13) is true.
Case 1: x= 0. This is impossible by Lemma 4.
Case 2: y= 0. By Lemma 5, it follows that
(n, x, y, z) = (2,3,0,3).
Case 3: x≥1and y≥1. Assume that xis
even. It follows that x= 2u, for some positive
integer u. Since n≡2 (mod 30), we obtain that
n≡2 (mod 3), and so nx≡2x≡4u≡1 (mod 3).
Then z2=nx+ 10y≡2 (mod 3). This is im-
possible since z2≡0,1 (mod 3). Thus xis odd.
There exists a non-negative integer vsuch that
x= 2v+ 1. Since n≡2 (mod 30), we obtain that
nx=n2v+1 ≡22v+1 (mod 30).
Subcase 3.1: vis even. Then v= 2a, for some
non-negative integer a. Since 24a≡16a≡1
(mod 5), it follows that nx≡24a+1 ≡2 (mod 10),
and so z2=nx+ 10y≡2 (mod10). This is
impossible by Theorem 1.
Subcase 3.2: vis odd. Then there exists a non-
negative integer bsuch that v= 2b+ 1. Since 24b≡
16b≡1 (mod 5), it follows that nx≡24b+3 ≡
8 (mod 10), and so z2=nx+ 10y≡8 (mod 10).
This is impossible by Theorem 1.
By Theorem 7, we have the following examples
and the corollary.
Example 8. The Diophantine equation 2x+10y=z2
has a unique non-negative integer solution. The so-
lution is (x, y, z) = (3,0,3).
Example 9. The Diophantine equation 32x+ 10y
=z2has no non-negative integer solution.
Corollary 10. Let mand nbe positive integers with
n≡2 (mod 30). Then the Diophantine equation
nx+ 10y=z2m(14)
has a unique non-negative integer solution. The solu-
tion is (n, m, x, y, z) = (2,1,3,0,3).
Proof. Let a, b and cbe non-negative integers such
that the equation (14) is true. Therefore (x, y, z) =
(a, b, cm)is a solution of the equation (13). By The-
orem 7, we get n= 2, a = 3, b = 0 and cm= 3.
Then c= 3 and m= 1. Hence (n, m, x, y, z) =
(2,1,3,0,3) is the only one solution of the equation
(14).
Theorem 11. Let nbe prime with n≥7,n≡
1 (mod 4) and n≡ 1 (mod 5). If yis even, then the
Diophantine equation (13) has no non-negative inte-
ger solution.
Proof. Let x, y and zbe non-negative integers such
that the equation (13) is true. Since yis even, we
have y= 2k, for some non-negative integer k.
Case 1: k= 0. Then y= 0. From (13), we have
z2−nx= 1.(15)
It is easy to check that z > 1and x > 0. Assume that
x > 1. Then min{z, n, 2, x}>1. By Theorem 3 and
(15), we have n= 2, a contradiction. Thus x= 1,
and so (z−1)(z+ 1) = n. Since nis prime, we get
z−1=1and z+ 1 = n. Thus z= 2 and n= 3, a
contradiction.
Case 2: k > 0. From (13), it follows that
(z−10k)(z+ 10k) = nx.(16)
Since nis prime, there exists a non-negative integer
hsuch that
z−10k=nh(17)
and
z+ 10k=nx−h.(18)
From (17) and (18), we get x > 2hand
2·10k=nh(nx−2h−1).(19)
Since nis prime with n≥7, we obtain that h= 0
and
2·10k=nx−1 = (n−1)(nx−1+nx−2+· · · + 1).
Since k > 1and n−1>2, it follows that 4|(n−1) or
5|(n−1). Thus n≡1 (mod 4) or n≡1 (mod 5),
a contradiction.
Corollary 12. The Diophantine equation
7x+ 100y=z2(20)
has no non-negative integer solution.
Proof. Assume that (a, b, c)is a non-negative integer
solution of (20). Therefore 7a+ 102b=c2. Thus
(n, x, y, z) = (7, a, 2b, c)is a non-negative integer
solution of (13). This is impossible by Theorem 11.
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Acknowledgment
The author would like to thank reviewers for careful
reading of this manuscript and the useful comments.
This work was supported by Research and Develop-
ment Institute and Faculty of Science and Technol-
ogy, Thepsatri Rajabhat University, Thailand.
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This work was supported by Research and Develop-
ment Institute and Faculty of Science and Technol-
ogy, Thepsatri Rajabhat University, Thailand.
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