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k a March Highlights and 2025 AoPS Online Class Information
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Mar 2, 2025
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jlacosta
Mar 2, 2025
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Peter   4
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sansgankrsngupta
Mar 13, 2025
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May 25, 2007
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May 25, 2007
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C 2
Peter   6
N May 24, 2022 by ZETA_in_olympiad
The positive integers $a$ and $b$ are such that the numbers $15a+16b$ and $16a-15b$ are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares?
6 replies
Peter
May 25, 2007
ZETA_in_olympiad
May 24, 2022
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Peter
3615 posts
#1 • 2 Y
Y by Adventure10, UpvoteFarm
The positive integers $a$ and $b$ are such that the numbers $15a+16b$ and $16a-15b$ are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares?
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mathmanman
1444 posts
#2 • 2 Y
Y by Adventure10, UpvoteFarm
The answer is $231361$ :
$a = 14911,$
$b = 481.$

$15 \cdot 14911 + 16 \cdot 481 = 231361 = 481^2,$
$16 \cdot 14911 - 15 \cdot 481 = 231361 = 481^2.$

Now, let us prove it.
We have :
$E_1 : 15a + 16b = x^2,$
$E_2 : 16a - 15b = y^2.$

$E_1^2 + E_2^2 \implies 481(a^2+b^2) = x^4 + y^4.$

So $x^4 + y^4 = 0 \mod 13 \cdot 37.$

But, this implies $x^4 + y^4 = 0 \mod 13$ and $x^4 + y^4 = 0 \mod 37$ that is, since these are primes, and taking $x$ and $y$ to be non-zero in $\mathbb{Z}/13\mathbb{Z}$ and $\mathbb{Z}/37\mathbb{Z}$ :
$\left(\frac xy \right)^4 = -1 \mod 13$ and $\left(\frac xy \right)^4 = -1 \mod 37.$

But, this is impossible, which implies that $x$ and $y$ are both divisible by $13$ and $37,$ hence $x = 481u$ and $u = 481v,$ hence $a^2 + b^2 = 481^3(u^2+v^2).$

The possible minimum for $\min(u, v)$ is $1.$
But, $1^2 + 31^2 = 2 \cdot 481 \implies 481^2 + (31 \cdot 481)^2 = 481^3(1^2 + 1^2).$

And, (today's a lucky day :) ), this necessary condition happens to be sufficient, giving the result announced in the very beginning of this post.
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ideahitme
91 posts
#3 • 2 Y
Y by Adventure10, UpvoteFarm
Peter wrote:
The positive integers $ a$ and $ b$ are such that the numbers $ 15a + 16b$ and $ 16a - 15b$ are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares?

We characterize the set
\[ \mathcal{S} = \{\; (a,b)\;\vert\; 15a + 16b = x^{2},\; 16a - 15b = y^{2}\;\text{for some}\; x,y\in\mathbb{N }\;\}.
\]
Let $ (a,b)\in\mathcal{S}$. We can find positive integers $ x$ and $ y$ such that $ 15a + 16b = x^{2}$ and $ 16a - 15b = y^{2}$. After solving the system of equations $ 15a + 16b = x^{2}$, $ 16a - 15b = y^{2}$, we obtain
\[ a = \frac { 15x^{2} + 16y^{2}}{15^{2} + 16^{2}} = \frac { 15x^{2} + 16y^{2}}{13\cdot 17},\; b = \frac {16x^{2} - 15y^{2}}{15^{2} + 16^{2}} = \frac {16x^{2} - 15y^{2}}{13\cdot 17}.
\]
Now, we have to work modulo $ 13$ and modulo $ 17$.

Step 1. First, we work modulo $ 13$. We need to solve the system of congruences
\[ 15x^{2} + 16y^{2}\equiv 0\; (mod\; 13)\;\;\text{and}\;\; 16x^{2} - 15y^{2}\equiv 0\; (mod\; 13).
\]
The first equation reads $ 2x^{2} + 4y^{2}\equiv 0\; (mod\; 13)$ or $ x^{2}\equiv - 2y^{2}\; (mod\; 13)$. The second equation reads $ 3x^{2} - 2y^{2}\equiv 0\; (mod\; 13)$ or $ 3x^{2}\equiv 2y^{2}\; (mod\; 13)$. We apply Gauss Elimination. We compute $ 3 ( - 2y^{2})\equiv 2y^{2}\; (mod\; 13)$ or $ 8y^{2}\equiv 0\; (mod\; 13)$. This implies that $ y\equiv 0\; (mod\; 13)$. Also, we find that $ x^{2}\equiv - 2y^{2}\equiv 0\; (mod\; 13)$ so that $ x\equiv 0\; (mod\; 13)$. We conclude that $ x\equiv y\equiv 0\; (mod\; 13)$ is the unique solution of the system of congruences.

\textsc{Step 2.} Second, we work modulo $ 17$. We need to solve the system of congruences
\[ 15x^{2} + 16y^{2}\equiv 0\; (mod\; 17)\;\;\text{and}\;\; 16x^{2} - 15y^{2}\equiv 0\; (mod\; 17).
\]
The first equation reads $ - 2x^{2} - y^{2}\equiv 0\; (mod\; 17)$ or $ 2x^{2}\equiv y^{2}\; (mod\; 17)$. The second equation reads $ - x^{2} + 2y^{2}\equiv 0\; (mod\; 17)$ or $ x^{2}\equiv 2y^{2}\; (mod\; 17)$. Hence, we compute $ 2 ( 2y^{2})\equiv y^{2}\; (mod\; 17)$ or $ 4y^{2}\equiv 0\; (mod\; 17)$. This implies that $ y\equiv 0\; (mod\; 17)$. Also, we find that $ x^{2}\equiv 2y^{2}\equiv 0\; (mod\; 17)$ so that $ x\equiv 0\; (mod\; 17)$. We conclude that $ x\equiv y\equiv 0\; (mod\; 17)$ is the unique solution of the system of congruences.

Combining results from Step 1 and Step 2, we can write $ x = 13\cdot 17 u$ and $ y = 13\cdot 17 v$ for some positive integers $ u$ and $ v$. Now, observe that
\[ 15a + 16b = x^{2},\; 16a - 15b = y^{2}
\]
becomes
\[ a = \frac { 15x^{2} + 16y^{2}}{13\cdot 17},\; b = \frac {16x^{2} - 15y^{2}}{13\cdot 17}
\]
or
\[ a = 13\cdot 17 (15u^{2} + 16v^{2}) ,\; b = 13\cdot 17 (16u^{2} - 15v^{2}).
\]
Hence, it follows that the set $ \mathcal{S}$ can be represented as
\[ \mathcal{S} = \{\; (a,b)\;\vert\; a = 13\cdot 17 (15u^{2} + 16v^{2}) ,\; b = 13\cdot 17 (16u^{2} - 15v^{2}), u,v\in\mathbb{N }\;\}.
\]
We therefore conclude that the minimum value is $ \left( 13\cdot 17\right)^{2}$.

Notes

In the solution, we met the systems of congruences. We recall the congruences from Step 1: $ 15x^{2} + 16y^{2}\equiv 0\; (mod\; 13)$, $ 16x^{2} - 15y^{2}\equiv 0\; (mod\; 13)$. There are many alternative ways to reach $ x\equiv y\equiv 0\; (mod\; 13)$.

Method 1. One may use Brahmagupta-Fibonacci Identity:
\[ \left( A^{2} + B^{2}\right)\left( X^{2} + Y^{2}\right) = \left( AX + BY\right)^{2} + \left( AY - BX\right)^{2}.
\]
Indeed, we obtain
\[ \left({15}^{2} + {16}^{2}\right)\left( a^{2} + b^{2}\right) = \left( 15a + 16b\right)^{2} + \left( 16a - 15b\right)^{2} = x^{4} + y^{4}.
\]
Since $ {15}^{2} + {16}^{2} = 13\cdot 17$, we find that $ x^{4} + y^{4}$ is divisibly by $ 13$. We now apply the following result with $ p = 13$ to conclude that $ x\equiv y\equiv 0\; (mod\; 13)$.

Proposition. Let $ p$ be a prime with $ p\equiv 5\; (mod\; 8)$. Suppose that $ a^{4} + b^{4}$ is divisible by $ p$ for some integers $ a$ and $ b$. Then, both $ a$ and $ b$ are divisible by $ p$.


Proof. Assume to the contrary that at least one of them are not divisible by $ p$. Since $ p$ divides $ a^{4} + b^{4}$, we see that none of them are divisible by $ p$. Fermat's Little Theorem yields that $ a^{p - 1}\equiv b^{p - 1}\equiv 1\; (mod\; p)$. Since $ p\equiv 5\; (mod\;8)$ or since $ \frac {p - 1}{4}$ is odd, we have $ ( - 1)^{\frac {p - 1}{4}} = - 1$. Since $ p$ divides $ a^{4} + b^{4}$, we obtain
\[ a^{4}\equiv - b^{4}\; (mod\; p).
\]
Raise both sides of the congruence to the power $ \frac {p - 1}{4}$ to obtain
\[ a^{p - 1}\equiv ( - 1)^{\frac {p - 1}{4}}b^{p - 1}\equiv - b^{p - 1}\; (mod\; p).
\]
or
\[ 1\equiv a^{p - 1}\equiv - b^{p - 1}\equiv - 1\; (mod\; p).
\]
This is a contradiction because $ p$ is an odd prime. Therefore, both $ a$ and $ b$ are divisible by $ p$.

Method 2. We now work on the field $ \mathbb{Z}/{13\mathbb{Z}}$. (Since $ 13$ is prime, we know that the ring $ \mathbb{Z}/{13\mathbb{Z}}$ is a field. We identify $ \overline{\alpha}\in\mathbb{Z}/{13\mathbb{Z}}$ with $ \alpha$.) The above congruences become
\[ 2x^{2} + 3y^{2} = 0\;\;\text{and}\;\; 3x^{2} - 2y^{2} = 0.
\]
Of course, Gauss Elimination works. However, we offer an alternative way. Observe that
\[ \left[\begin{array}{cc}x^{2} & y^{2} \\
- y^{2} & x^{2}\end{array}\right]\left[\begin{array}{c}2 \\
3\end{array}\right] = \left[\begin{array}{c}0 \\
0\end{array}\right].
\]
This implies that $ \left[\begin{array}{cc}x^{2} & y^{2} \\
- y^{2} & x^{2}\end{array}\right]$ has zero determinant so that $ x^{4} + y^{4} = 0$. We compute
\[ x^{12} = \left( x^{4}\right)^{3} = \left( - y^{4}\right)^{3} = - y^{12}
\]
If $ y\neq 0$, then we also have $ x = 0$. Hence, we obtain $ 1 = x^{12} = - y^{12} = - 1$, which is a contradiction. Therefore, we get $ y = 0$ and so $ x = 0$.
This post has been edited 1 time. Last edited by ideahitme, Aug 20, 2008, 3:33 PM
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Chan-Nor
2 posts
#4 • 3 Y
Y by Adventure10, UpvoteFarm, Mango247
Hi Ideahitme,

I'am sorry to say this but it seems to me that there is something wrong in your solution

Namely,



15^2 + 16^2 = 481

but

481 = 13 x 37

( and not as you write 13 x 17)

Thus we work modulo 13 and 37.

So, in Step 2 we consider


15x^2 + 16 y ^2 = 0 mod 37
and
16x^2 - 15y^2 = 0 mod 37


:lol: :wink:
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Peter
3615 posts
#5 • 2 Y
Y by Adventure10, UpvoteFarm
mathmanman wrote:
$ \left(\frac xy \right)^4 = - 1 \mod 13$ and $ \left(\frac xy \right)^4 = - 1 \mod 37.$
It took me really long to realize these were fractions and not L\'egendre symbols.

Just posting this for the next person.
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scorpius119
1677 posts
#6 • 3 Y
Y by UpvoteFarm, Adventure10, Mango247
\[ 15a + 16b = x^2,16a - 15b = y^2\Rightarrow (15^2 + 16^2)a = 15x^2 + 16y^2
\]
Now $ 15^2 + 16^2 = 481 = 13\cdot 37$ so we need the congruences
\[ 15x^2 + 16y^2\equiv 0\pmod{p}
\]
for each of $ p = 13,37$. In each case, if $ p$ divides $ x$, then $ p$ divides $ y$ as well. Otherwise, we get
\[ (4yx^{ - 1})^2\equiv - 15\pmod{p}
\]
We can check (using QR for example, or the long way... :roll: ) that -15 is not a square modulo either prime, so we necessarily find that 481 divides both $ x,y$. But in this case, if $ x = 481m,y = 481n$, then there is always an integer solution for $ a,b$, namely
\[ a = 481(15m^2 + 16n^2),b = 481(16m^2 - 15n^2)
\]
In particular, there's a positive integer solution for $ m = n = 1$, that is, $ x = y = 481$, for $ a = 31\cdot 481,b = 481$. Since $ x,y$ positive and divisible by 481, the minimum is $ \boxed{481}$.
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ZETA_in_olympiad
2211 posts
#7 • 1 Y
Y by UpvoteFarm
Let $15a+16b=m^2$ and $16a-15b=n^2.$
Sum up their squares to get $$m^4+n^4=481(a^2+b^2)=13\cdot 37(a^2+b^2).$$So $13, 37 \mid m^4+n^4.$

Claim: $13, 37\mid m+n.$
Proof. Assume the contrary that $13\nmid m+n.$ The case with 37 is similar. So we have $$m^4\equiv -n^4\pmod{13} \implies m^{12}\equiv- n^{12} \pmod{13}.$$By Fermat's we get the contradiction that $2\equiv 0 \pmod{13}.$ $\blacksquare$

Let $m=481x, n=481y.$ So $$a=481(15x^2+16y^2) \text{ and } b=481 (16x^2-15y^2).$$For the least, set $x=y=1,$ to get $$a=14911 \text{ and }b=481 \implies b^2=231361.$$
This post has been edited 1 time. Last edited by ZETA_in_olympiad, May 24, 2022, 10:07 AM
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