# Tag Info

### $a^2+a+3b$ 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
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

• 12.1k
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

### 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

### $\lfloor x^n\rfloor\lfloor y^n\rfloor$ 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

• 106k
Accepted

• 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

Only top scored, non community-wiki answers of a minimum length are eligible