## Proving an expression is perfect square

I have this expression I got in one larger exercise: (2+√3)2n+1+(2−√3)2n+1−46(2+√3)2(n+1)+1+(2−√3)2(n+1)+1−46+1 and i need to prove it is perfect square. I tried many different approaches but couldn’t find way to show it is square. Interesting fact is (2+√3)(2−√3)=1 so I tried replacing (2+√3)=x and (2−√3)=1/x to see if I would get an idea. Alternative form … Read more

## Is $67500y^3-y$ ever a square?

Is there an integer $y \neq 0$ where $67500y^3-y$ is a square? So far I’ve tried a computer search with no luck. Answer No. Let $y$ be an integer and assume $67500y^3-y = y(67500y^2-1)$ is a square. Since for any prime $p$ (EDIT: which divides $y$), the $p$-adic valuation $val_p(67500y^3-y) = val_p(y)$, $y$ has to … Read more

## Can a 300-digit number whose digits are 100 0s, 100 1s and 100 2s (in some order) be a perfect square?

The question: Can a 300-digit number whose digits are 100 0s, 100 1s and 100 2s (in some order) be a perfect square? Prove your answer. I’m new to digital roots in divisibility and I am trying to figure this out. Am I right in saying that the digital roots for this would be: 100(0)+100(1)+100(2)=300100(0+1+2)=300→3+0+0=3 … Read more

## Arrange nn tiles in a rectangle pattern

Is there a way to find how to arrange n tiles in order to form a rectangle that is closest to a square? As I result I am looking for the dimension of that rectangle if it exists. For example, here are the results I expect from 1 to 9. a 1 by 1 square, … Read more

## Given that n4−4n3+14n2−20n+10n^4-4n^3+14n^2-20n+10 is a perfect square, find all integers n that satisfy the condition

So, I tried solving that by n4−4n3+14n2−20n+10=x210=x2−a2,a2=n4−4n3+14n2−20n+1010=(x+a)(x−a) but I couldn’t find any integers when I solved it Answer It always helps to form squares from the biggest power and is a good strategy: n4−4n3+14n2−20n+10=n2(n2−4n+4)+10n2−20n+10=n2(n−2)2+10(n−1)2=(n(n−2))2+10(n−1)2=((n−1)2−1)2+10(n−1)2=(n−1)4−2(n−1)2+1+10(n−1)2=(n−1)4+8(n−1)2+1=((n−1)2+4)2−15=x2 I think this quite large hint will make it a bit easier to solve it. Just as a note: it just … Read more

## A slight variation on the Pythagorean theorem

Are there any solution to a^2+b^2=c^2+1 where a \not=0 and b\not=0 This is a follow-on from a previous question For what n and m is this number a perfect square?, which ultimately boils down to the above. Any help would be greatly appreciated. Answer More generally, the complete answer to, x_1^2+x_2^2=y_1^2+y_2^2 was known way back … Read more

## Supposed $a,b \in \mathbb{Z}$. If $ab$ is odd, then $a^{2} + b^{2}$ is even.

Supposed $a,b \in \mathbb{Z}$. If $ab$ is odd, then $a^{2} + b^{2}$ is even. I’m stuck on the best way to get this started. My thinking is that I could use cases. i.e. Case 1: a is even and b is odd Case 2: a is odd and b is even Case 3: a is … Read more

## Are there no even squares expressible as the sum of two prime squares?

When I was playing around with different number sequences, I noticed that I couldn’t find any even squares that are expressible as the sum of two prime squares. Is this true, and is this related to Fermat’s theorem on sums of two squares? Answer Hint: Even squares are of necessity multiples of four (why ?). … Read more

## On a conjecture and other numbers like $144$.

Yesterday, I noticed that the following numbers were prime: $$67, 167, 467, 967, 1667.$$ I realised that the digits of these primes, apart from $67$, were square numbers: $$0^2, 1^2, 2^2, 3^2, 4^2.$$ I then wanted to find values $a_n$ such that when we concatenate $a_n^{\ \ 2}$ with $67$, it is a prime number. … Read more

## a,ba, b are the integer part and decimal fraction of √7\sqrt7 find integer part of ab\frac{a}{b}

a,b are the integer part and decimal fraction of √7 find integer part of ab using calculator : √7 = 2.645 ab=20.6=206=3.333 integer part of ab=3 how to do it without calculator? if i don’t know √7 part Answer Note that 2.5<√7<3 ab=2√7−2=2(√7+2)3 3=2(2.5+2)3<2(√7+2)3<2(3+2)3<4 Thus the integer part of 2√7−2 is 3 AttributionSource : Link … Read more