9
votes
$a^2+a+3b$ and $b^2+b+3a$ both perfect square
You have the right general idea, but the following shows how to use perfect square lower and upper limits to determine the solution. First, due to symmetry, WLOG consider that $a \ge b$. Next, if ...
- 45.8k
4
votes
Accepted
$a^2+a+3b$ and $b^2+b+3a$ both perfect square
You made a good start. Here we set aside the trivial solution $a=b=0$. All you need to do next is check through a few possibilities for $m$ and $n$; most potential values for $m$ and $n$ can be ruled ...
- 17.6k
4
votes
Solve the equation $3x^2+2y^2-4xy+x-2=0$ in the set of integer numbers.
Solving the quadratic equation $3x^2+(1-4y)x+2y^2-2=0$ in $x$, we obtain the discriminant
$$
D=-8y^2-8y+25
$$
which needs to be a square. Hence in particular $D\ge 0$. This is equivalent to $y=1,0,-1,-...
- 128k
3
votes
Accepted
Solve the equation $3x^2+2y^2-4xy+x-2=0$ in the set of integer numbers.
You have
$$2(x - y)^2 = 2 - x^2 - x \implies (x - y)^2 = 1 - \dfrac{x(x+1)}{2}.$$
This implies that $1 - x(x+1)/2$ is a perfect square. Since the term $x(x+1)/2$ is always non-negative, then either $(...
- 12.1k
3
votes
Accepted
Show that if the numbers $\sqrt{a^2+b+c+1}$, $\sqrt{b^2+c+a+1}$ and $\sqrt{c^2+a+b+1 }$ are rational, then $a = b = c$
It's clear that the numbers in the radical must be perfect squares，you could choose the largest number of $a,b,c$, WLOG, assume $a \ge b \ge c$.
Notice that $$a^2 \lt a^2+b+c+1 \le (a+1)^2$$
you only ...
- 98
3
votes
How to find values of the given $4n^2 - 5n + 16$ for which the function is a perfect square?
Complete the square on the right hand side to get $m^2=(2n-5/4)^2+231/16$, or clearing denominators:
$$(4m)^2=(8n-5)^2+231$$
You can then write $231$ as the difference of two squares, and find all ...
- 542
2
votes
$\lfloor x^n\rfloor\lfloor y^n\rfloor$ is a perfect square
If the conditions are slightly weakened in either of two ways, or ceiling is used instead of floor, the answer is "no". This is because of the existence of non-integral Pisot–Vijayaraghavan ...
- 639
2
votes
Solve the equation $3x^2+2y^2-4xy+x-2=0$ in the set of integer numbers.
$3x^2 + 2y^2 - 4xy + x - 2 = 2(x - y)^2 + (x + 2)(x - 1) = 0$
$2(x - y)^2 = 0$ AND $(x + 2)(x - 1) = 0$
$x = y$ AND $x = -2$ or $x = 1$
So, $ y = -2$ or $y = 1$
For the other solutions, solve:
$2(x - ...
- 434
2
votes
Solve the equation $3x^2+2y^2-4xy+x-2=0$ in the set of integer numbers.
$3x^2+2y^2-4xy+x-2=0
$
$2y^2-4xy+3x^2+x-2=0
$
$\begin{array}\\
d
&=16x^2-4\cdot 2(3x^2+x-2)\\
&=16x^2-24x^2-8x+16\\
&=-8x^2-8x+16\\
&=-8(x^2+x-2)\\
&=-8(x+2)(x-1)\\
&=-2(4x^2+...
- 106k
2
votes
Accepted
Others abbreviated calculation formulas...
You can if you are ok with complex numbers. For example,
$$x^2+y^2+z^2 = \left(\sqrt{x^2+y^2}+iz\right)\left(\sqrt{x^2+y^2}-iz\right).$$
For the second problem:
$$a^2+b^2+c^2+2a+2b+2c+3$$
$$=a(a+2+1/...
- 1,750
1
vote
Number theory on multiples of squares
If you add $2^2\cdot 7^2\cdot 5^2=4900$ to your solution, you get another $4948,4949,4950$ that satisfy the same property (with the same squares in fact). You can do this as many times as you like, so ...
- 4,714
1
vote
Show that if the numbers $\sqrt{a^2+b+c+1}$, $\sqrt{b^2+c+a+1}$ and $\sqrt{c^2+a+b+1 }$ are rational, then $a = b = c$
In order for the expression inside the first root to be a perfect square, you need that
$$a^2+b+c+1 = (a+x)^2$$ for some $x$.
The factors of $1$ are $\pm 1$. Thus,
$$a^2+b+c+1 = (\pm a \pm 1)^2.$$
If ...
- 1,750
1
vote
Show that if the numbers $\sqrt{a^2+b+c+1}$, $\sqrt{b^2+c+a+1}$ and $\sqrt{c^2+a+b+1 }$ are rational, then $a = b = c$
Hint: Assume, WLOG, $a\geq b\geq c$ and consider the implications of $a^2+b+c+1$ being a perfect square.
- 31.4k
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