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Today I've got a math contest. And I couldn't solve the following problem :

Let $m$ be a natural number, where : $m+7$ is a perfect square & $m-34$ is also a perfect square .

What's the value of $m$ ?

I have absolutely no idea on how should I proceed with this kind of problems and I'm completly stuck at it. I don't even know how to start .

DeltaWeb
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  • Did you literally not know how to start? As in, you hadn't gotten as far as transcribing the statement of the problem into $m+7 = x^2, m-34 = y^2$? There's often a wide degree of interpretation in "don't know how to start", so it's helpful if you can try to provide more detail about what you tried to do (even if it seems like nothing!) and what "this kind of problems" means. – Erick Wong May 07 '16 at 15:46

3 Answers3

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Let $p^2=m+7$, $q^2=m-34$.

We have:

$$p^2-q^2=41$$

Try continuing form here.

Kenny Lau
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(Another approach — Just for fun!)

Observe that the squares are precisely the sums of odds:

$1 = 1^2$ (first $1$ odd)

$1+3 = 4 = 2^2$ (first $2$ odds)

$1+3+5 = 9 = 3^2$ (first $3$ odds)

$1+3+5+7 = 16 = 4^2$ (first $4$ odds), and so forth.

You have two numbers with difference $41$ as squares; note that $41$ is the $21$st odd.

So the smaller is the sum of the first $20$ odds, and the latter is the sum of the first $21$ odds.

The sum of the first $n$ odds is $n^2$, thus, the first $20$ odds sum to $20^2 = 400$.

You are told that $m - 34 = 400$; so $m = 434$.

Benjamin Dickman
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Another possibility is to see that the difference between $n^2$ and $(n+1)^2$ is $2n+1$. With a $34+7=41=2n+1$ gap, we have a solution given by $n=20$, giving

$n^2=400 < (n+1)^2=441$ thus $m=400+34=441-7=434$;

but it could happen that such a gap occurs between non-consecutive squares.

If such is the case, as the gaps are larger and larger, this gap should take place before $20^2=400$. But there is no solution because if $a^2-b^2=41 \Leftrightarrow (a-b)(a+b)=41$; as 41 is a prime number, necessarily $a-b=1$ i.e., a and b differ by one unity, and we are back to the previous (and unique) solution.

Jean Marie
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