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Question: Find all integer values of $x$ such that $x^2 + 13x + 3$ is a perfect integer square.

What I have attempted;

For $x^2 + 13x + 3$ to be a perfect integer square let it equal $k^2$ where $k \in \mathbb{Z} $

Hence $$x^2 + 13x + 3 = k^2$$

$$ \Leftrightarrow x^2 + 13x + (3-k^2)=0$$

$$ \Leftrightarrow x={-b\pm\sqrt{b^2-4ac} \over 2a} $$

$$ \Leftrightarrow x={-13\pm\sqrt{13^2-4(1)(3-k^2)} \over 2(1)}$$

$$ \Leftrightarrow x={-13\pm\sqrt{169-12+4k^2} \over 2}$$

$$ \Leftrightarrow x={-13\pm\sqrt{157+4k^2} \over 2}$$

We also want $157+4k^2$ to be a perfect square hence

$$ 157 + 4k^2 = n^2 $$

$$ n^2 -4k^2 = 157 $$

$$ (n-2k)(n+2k) = 157 $$

What should I do now?

I have looked at this; Find all integers $x$ such that $x^2+3x+24$ is a perfect square.

but I am confused what I should do to continue..

Community
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bigfocalchord
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    Almost there. How can you write 157 as the product of two integers? – N. S. May 05 '16 at 04:16
  • $157 = ab,\ a = n!-!2k,\ b = n!+!2k !\iff!$ $ n = (b!+!a)/2,\ k = (b!-!a)/4,,$ which are both integers $\iff! b\equiv a\pmod{!4}.,$ So find all splittings of $157$ into factors congruent $!\bmod 4,$ (easy by $157$ is prime). $\ \ $ – Bill Dubuque Mar 16 '25 at 20:35
  • I solved your question completely with considering all answers and cases ! Cause 157 is prime there exists just four cases and with checking each case we have an answer. Note that $157=(-157)(-1)=(157)(1)$ – Benyamin Khanzadeh H. Mar 17 '25 at 07:55

2 Answers2

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Observe that $x^2+13x+3=(x+\frac{13}{2})^2-\frac{157}{4}$

As you have assumed, let $(x+\frac{13}{2})^2-\frac{157}{4}=k^2$

Hence, by simplifying, we get that $(2x+13)^2-157=4k^2$

Further simplifying by factorisation yields $(2x+13-2k)(2x+13+2k)=157=157\cdot 1$

So, there are only two possibilities:

  1. $(2x+13-2k)=157$ and $(2x+13+2k)=1$
  2. $(2x+13-2k)=1$ and $(2x+13+2k)=157$

The first one yields $k=-39$ and the second one yields $k=39$.

Can you find $x$ now??

SchrodingersCat
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As something you did:

For $x^2 + 13x + 3$ to be a perfect integer square let it equal $k^2$ where $k \in \mathbb{Z} $ Hence $$x^2 + 13x + 3 = k^2$$ $$ \Leftrightarrow x^2 + 13x + (3-k^2)=0$$ $$ \Leftrightarrow x={-b\pm\sqrt{b^2-4ac} \over 2a} $$ $$ \Leftrightarrow x={-13\pm\sqrt{13^2-4(1)(3-k^2)} \over 2(1)}$$ $$ \Leftrightarrow x={-13\pm\sqrt{169-12+4k^2} \over 2}$$ $$ \Leftrightarrow x={-13\pm\sqrt{157+4k^2} \over 2}$$ We also want $157+4k^2$ to be a perfect square hence $$ 157 + 4k^2 = n^2 $$ $$ n^2 -4k^2 = 157 $$ $$ (n-2k)(n+2k) = 157 $$

157 is a prime number! and $(n-2k)$ and $(n+2k)$ both are integers, so we have 4 cases:

Case 1. $n-2k = 1 , n+2k = 157$

Hence: $ n = 79, , k=39$

Case 2. $n-2k = -1 , n+2k = -157$

Hence: $ n = -79, , k=-39$

Case 3. $n-2k = -157 , n+2k = -1$

Hence: $ n = -79, , k=39$

Case 4. $n-2k = 157 , n+2k = 1$

Hence: $ n = 79, , k=-39$

So we got: $(n,k)\in \{(79,39),(-79,39),(79,-39),(-79,-39)\}$

Benyamin Khanzadeh H.
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